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The definite integral as area
A plot of land beside a road. Three sides are straight, and the far side follows the curved bank of a canal.
Picture a plot of land beside a road. One side runs straight along the road. The two ends are straight too, at right angles to the road. The far side follows the bank of a canal, and the bank curves.
the far side drawn straight
the real bank
A straight far side makes a trapezium, area 9.375. The real bank curves: no Class 10 formula gives its area.
The land record needs the area of this plot. By Class 10 you knew the area of a rectangle, a triangle, a trapezium and a circle. None of those formulas fits a curved side.
Chapter 7’s way: cut the plot into strips and add them. Six strips give about 7.34; F(4) − F(1) gives the limit, 8.25.
Chapter 7 met this problem once already. It cut the area under a graph into thin strips and added them. Then the fundamental theorem turned that long total into one subtraction, $F(b) - F(a)$.
A circle, an ellipse, and a parabola cut off by a line. Each region has a curved edge. Can one subtraction find all three?
Naming the open question
Here is the question to carry to the last page. Can that one subtraction find the area of any region with a curved edge? Think of a circle. You were told its formula but never shown where it comes from. Think of an ellipse, or a piece of a parabola. You were never given a formula for those.
y equals x squared over 4 plus 1, above the axis from x equals 1 to 4. The integral and the area are both 8.25.
Start with the plot. Put the road on the $x$-axis and the two ends at $x = 1$ and $x = 4$. Let the canal bank be the curve $y = x^2/4 + 1$. The figure shades the plot. Its area is $\int_1^4 (x^2/4 + 1) \mathrm{d} x = 8.25$ square units.
The plot comes back several times. Keep it in mind.
A SCHOLAR INDIA REMEMBERS
Aryabhata finished the Aryabhatiya at Kusumapura, near modern Patna, in 499 CE, when he was $23$. His methods rested on reasoning anyone could check. Check the plot before you trust $8.25$. It is $3$ wide. The bank is $1.25$ high at the left end and rises to $5$ at the right. So the area is more than $3 \times 1.25 = 3.75$ and less than $3 \times 5 = 15$. And $8.25$ sits between them.
Aryabhata · the hoopoe
Check yourself
Before any extra condition on $f$ is added, what does $\int_a^b f(x) d x$ compute?
the slope of the curve at $x = a$
a single number
the number of times the curve crosses the x-axis
the area under the curve, always
Check your answer
the slope of the curve at $x = a$ — Slope comes from differentiating $f$, not from integrating it. The definite integral accumulates; it does not measure a rate of change.
✓ a single number — (B) The definite integral always computes one number. That number equals an area only once $f(x) \geq 0$ is added as a condition on $[a, b]$.
the number of times the curve crosses the x-axis — Root count is a separate fact about $f$, found by solving $f(x) = 0$. The integral is a computed number that has nothing to do with how many times the curve crosses the axis.
the area under the curve, always — The area reading needs $f(x) \geq 0$ on $[a, b]$ first. Without it, the integral is still a number, just not automatically an area.
Under what condition does $\int_a^b f(x) d x$ equal the area between $y = f(x)$, the x-axis, and $x = a$, $x = b$?
$a < b$
$f$ is continuous on $[a, b]$
$f(x) \geq 0$ on $[a, b]$
$f$ is increasing on $[a, b]$
Check your answer
$a < b$ — Ordering the bounds is a separate bookkeeping fact, needed before the integral is even set up. It does not guarantee the curve stays above the axis.
$f$ is continuous on $[a, b]$ — Continuity is what makes the integral exist in the first place. It does not stop $f$ from dipping below the axis, so it is not the condition for the area reading.
✓ $f(x) \geq 0$ on $[a, b]$ — (C) The area reading holds exactly when the curve never dips below the x-axis on $[a, b]$, so $f(x) \geq 0$ throughout.
$f$ is increasing on $[a, b]$ — A curve can increase the whole way from a negative value to a positive one, dipping below the axis for part of $[a, b]$. Increasing says nothing about sign.
If $f(x)$ is negative on part of $[a, b]$, how does $\int_a^b f(x) d x$ compare with the TRUE area of the region?
it is larger, because negative values are counted twice
it becomes undefined, since the area formula only applies when $f(x) \geq 0$
it stays equal, since integration ignores sign
it is smaller, because the negative part subtracts instead of adding
Check your answer
it is larger, because negative values are counted twice — A negative piece of the integral is counted once, with a minus sign, it is never doubled. That is exactly why it makes the total smaller, not larger.
it becomes undefined, since the area formula only applies when $f(x) \geq 0$ — The integral is still perfectly well defined for a negative $f$, only the direct area reading needs $f(x) \geq 0$. Compute the integral, then apply the sign-and-split rule to recover the true area.
it stays equal, since integration ignores sign — Integration is signed by nature, that is the entire reason area problems need a split-and-absolute-value step whenever the curve crosses the axis.
✓ it is smaller, because the negative part subtracts instead of adding — (D) Wherever $f(x) < 0$, that piece of the integral is negative and works against the positive pieces, so the raw integral comes out smaller than the true, all-positive area.
You met every idea below in an earlier class. Try each one below.
Closed interval (Class 11, Mathematics (NCERT Class 11), page 11). Here: Chapter 7’s fundamental theorem evaluates over a closed interval $[a,b]$. That is exactly the stretch between the two ordinates an area problem names first. Check: the closed interval $[a,b]=[1,4]$ spans $b-a=4-1=3$ between its two ordinates.
Area of a circle (Class 10, Mathematics (NCERT Class 10), page 155). Here: The familiar area formula for a circle is exactly the number this book now derives from one quarter-disc integral. It is no longer assumed. Check: a circle of radius $r=7$ has area $\pi r^2$, where $r^2=7^2=49$.
Sector minus triangle (Class 10, Mathematics (NCERT Class 10), page 156). Here: A segment’s area already subtracts one shape from another, a sector’s area minus a triangle’s area. Area between two curves subtracts top minus bottom, the identical move. Check: a quarter-circle sector of radius $2$ has area $(\pi \cdot 2^2)/4 = \pi$. The right triangle it contains has area $(1/2) \cdot 2 \cdot 2 = 2$. The segment left over is $\pi - 2$.
Equation of a circle (Class 11, Mathematics (NCERT Class 11), page 180). Here: Writing a circle’s equation with its centre at the origin gives the setup $x^2+y^2=a^2$. That is exactly where this book’s own circle example starts. Check: a circle centred at the origin with radius $a=3$ has equation $x^2+y^2=9$, since $a^2=3^2=9$.
Standard equation of a parabola (Class 11, Mathematics (NCERT Class 11), page 182). Here: A parabola’s standard equation, simplest at the origin, is exactly the $y^2=4 a x$ form this book’s parabola area problem starts from. Check: a parabola $y^2=4 a x$ with $a=5$ gives $y^2=20 x$, since $4 \cdot 5=20$.
Standard equation of an ellipse (Class 11, Mathematics (NCERT Class 11), page 188). Here: An ellipse’s standard equation, simplest at the origin, gives exactly the $x^2/a^2+y^2/b^2=1$ form this book’s ellipse area problem starts from. Check: an ellipse with $a=5$, $b=3$ has equation $x^2/25+y^2/9=1$, since $a^2=25$ and $b^2=9$.
If any of these felt new, read the page named before going on.
The same plot, marked into thin rows for sowing at right angles to the road. Each row runs from the road to the bank.
Go back to the plot. Say a farmer marks it into rows for sowing. Each row runs at right angles to the road, from the road to the bank. Make the rows very thin. Then each row is almost a rectangle. Its width is small, and its length is the distance from the road to the bank.
4 strips
8 strips
16 strips
the plot is 8.25
The plot as 4, 8 and 16 strips. The rectangles add to 6.91, 7.56 and 7.90, nearing 8.25.
So the area of the plot is the total of many thin rectangles. This is Chapter 7’s leaf on squared paper again, an uneven shape whose area was once estimated by counting unit squares, with strips in place of squares.
dA = y dx
One strip at x. Its width is dx and its height is f(x), the curve’s height there. Its area is dA = y dx.
Now name the parts. Take one strip at a point $x$ between $a$ and $b$. Its width is $\mathrm{d} x$. Its height is $y = f(x)$, the height of the curve there. This thin rectangle is the ELEMENTARY STRIP. Its area is the ELEMENTARY AREA, $\mathrm{d} A = y \mathrm{d} x$.
Under y equals 4 minus x squared over 5, from 0 to 4: one strip, four strips, then the whole region, about 11.73.
Add up the strips from $x = a$ to $x = b$. As the strips get thinner, the total becomes a definite integral.
$A = \int_a^b y \mathrm{d} x = \int_a^b f(x) \mathrm{d} x$.
One strip of the plot at x = 2. The bank is 2 above the road, the strip 0.1 wide: area 0.2.
Let us check one strip of the plot. At $x = 2$ the bank is at this height:
$y = 2^2/4 + 1 = 2$.
*A strip $0.1$ wide* there has area $\mathrm{d} A = y \mathrm{d} x$, so about:
$\mathrm{d} A \approx 2 \times 0.1 = 0.2$.
As the strips get thinner, their total gets closer and closer to the integral. The theorem from Chapter 7 works it out:
$\int_1^4 (x^2/4 + 1) \mathrm{d} x = [x^3/12 + x]_1^4$
$= (64/12 + 4) - (1/12 + 1)$
$= (64/12 - 1/12) + (4 - 1) = 63/12 + 3 = 8.25$.
That is the area of the plot, in square units.
one strip at x
→
height × width
f(x) × dx
→
add
from a to b
→
A = ∫ f(x) dx from a to b
Every area in one line: take one strip, multiply its height by its width, then add the strips from a to b.
So every area problem starts with one strip. Draw it first. It shows you the height to integrate. It also shows where the strips start and stop.
A SCHOLAR INDIA REMEMBERS
How close is close enough? Put a bound on the strips. Cut the plot into $4$ strips and take each one’s height at its left end. The bank rises, so every strip stays under it. They add to $6.91$, too small. Take the right ends instead, and every strip reaches above the bank: $9.73$, too big. With $16$ strips the bounds narrow to $7.90$ and $8.61$. The plot, $8.25$, lies between them.
Aryabhata · the hoopoe
Check yourself
In the elementary-strip picture, what are the width and height of the strip at $x$?
width $f(x)$, height $d x$
width $d x$, height $b - a$
width $d y$, height $x$
width $d x$, height $f(x)$
Check your answer
width $f(x)$, height $d x$ — The curve’s value $f(x)$ is the strip’s HEIGHT, not its width. The width is the tiny horizontal slice $d x$.
width $d x$, height $b - a$ — $b - a$ is the length of the WHOLE interval, not the height of one strip at $x$. Each strip’s height is the curve’s own value there, $f(x)$.
width $d y$, height $x$ — A width of $d y$ belongs to the y-axis version of this idea, where the strip is horizontal. The vertical strip used here has width $d x$.
✓ width $d x$, height $f(x)$ — (D) The strip sits at $x$, running a tiny width $d x$ across, with its height set by the curve itself, $y = f(x)$.
Why does integrating the strip area $y d x$ from $x = a$ to $x = b$ give the total area?
the formula only works at the single strip located at $x = (a+b)/2$
$d x$ is defined as the total area divided by the number of strips
the integral sums infinitely many thin strips across the interval
each strip already equals the total area on its own
Check your answer
the formula only works at the single strip located at $x = (a+b)/2$ — No single strip carries the total area, the integral sums every strip from $x = a$ to $x = b$, not just the one sitting at the midpoint.
$d x$ is defined as the total area divided by the number of strips — $d x$ is a small width chosen first, not a quantity worked out from the final area. The area is the OUTPUT of the sum, never an ingredient of it.
✓ the integral sums infinitely many thin strips across the interval — (C) The integral is exactly the limit of summing every strip’s tiny area, $y d x$, as $x$ sweeps from $a$ to $b$, the same Riemann-sum idea from Chapter 7, now read as area.
each strip already equals the total area on its own — A single strip’s area, $y d x$, is vanishingly small on its own. Only adding up ALL of them, across the whole interval, reaches the total area.
How does the elementary-strip picture relate to the Riemann-sum idea from Chapter 7?
it works only for straight-line curves, not curved ones
it replaces the Riemann sum with a completely different limiting process
it only applies once the curve has been proven continuous
it is the same summing-of-thin-slices idea, now read as area
Check your answer
it works only for straight-line curves, not curved ones — The elementary-strip idea applies to any curve the integral can handle, curved included, restricting it to straight lines has no basis in how the sum is built.
it replaces the Riemann sum with a completely different limiting process — No new limiting process is introduced. The elementary strip is exactly Chapter 7’s Riemann-sum rectangle, only now interpreted as area.
it only applies once the curve has been proven continuous — Continuity is enough for the integral to exist (Chapter 7); it is not a separate condition here.
✓ it is the same summing-of-thin-slices idea, now read as area — (D) Nothing new is added at the level of the sum itself, this chapter takes Chapter 7’s Riemann sum and gives it a geometric reading, area, instead of leaving it as an abstract limiting number.
Integrating the strip area $y d x$ from $x = a$ to $x = b$ produces which formula for the area under a curve?
$A = \int_a^b d x$, with $y$ left out entirely
$A = f(b) - f(a)$
$A = \int_a^b f(x) d x$, with $y = f(x)$
$A = (b - a) f(a)$
Check your answer
$A = \int_a^b d x$, with $y$ left out entirely — Without $y$, the integral just measures the interval’s length, $b - a$. The strip’s height, $y = f(x)$, has to stay in the formula.
$A = f(b) - f(a)$ — $f(b) - f(a)$ is how much $f$ itself changed, it says nothing about the strips’ areas being added up. The area comes from summing $y d x$, not from subtracting two function values.
✓ $A = \int_a^b f(x) d x$, with $y = f(x)$ — (C) Summing every strip’s area $y d x$ across the interval is exactly the formula $A = \int_a^b f(x) d x$, the strip picture and the area-under-a-curve formula are the same fact, seen two ways.
$A = (b - a) f(a)$ — $(b-a) f(a)$ is one flat rectangle using only the height at $x = a$. The actual formula sums every strip’s OWN height across the whole interval, not one estimate.
The region under the curve, above the x-axis, between the ordinates x = 1 and x = 4. For the plot it is 8.25.
Here is the rule the strips give. Let $f(x) \geq 0$ on $[a, b]$. Take the region between the curve $y = f(x)$, the $x$-axis and the lines $x = a$ and $x = b$. These two lines are the ORDINATES at $a$ and $b$. The area of the region is
$A = \int_a^b f(x) \mathrm{d} x$.
above the axis
the integral is the area
dips below
Left, the curve stays above the axis, so the integral is the area, 8.25. Right, it dips below, and the integral, 8/3, is not.
The condition matters. The curve must stay on or above the axis from $a$ to $b$. If part of it dips below, the integral is not the area. A split fixes that, a few pages on.
Flat strips against the y-axis: the plot read with flat strips, x = y²/4 + 1 from y = 1 to 4. Still 8.25.
Strips can also lie flat. Take a curve written as $x = g(y)$, with $g(y) \geq 0$. Cut the region between it, the $y$-axis and the lines $y = c$ and $y = d$ into flat strips. Each strip is $x$ long and $\mathrm{d} y$ thick. So the area is
$A = \int_c^d g(y) \mathrm{d} y$.
y equals x squared, 0 to 2. Upright strips: 8 over 3. Flat strips, from the curve to x equals 2: also 8 over 3.
The flat strips in pictures: the 2 by 4 rectangle is 8, the piece left of the curve 16/3, and 8 − 16/3 = 8/3.
Let us find one region both ways. Take the region under $y = x^2$ from $x = 0$ to $x = 2$. Work it with upright strips, then with flat ones.
Worked example
The area under $y = x^2$ from 0 to 2, both ways
$A = \int_0^2 x^2 \mathrm{d} x$ Upright strips: each is $x^2$ tall and $\mathrm{d} x$ wide, from $x = 0$ to $x = 2$.
$A = [x^3/3]_0^2$ Integrate $x^2$: add 1 to the power, and divide by the new power.
$A = 8/3 - 0 = 8/3$ Put in the top limit, then the bottom one, and subtract: this is $F(b) - F(a)$.
$x = \sqrt{y}$ For flat strips, solve $y = x^2$ for $x$: take the square root, and keep $x \geq 0$.
Strip length: $2 - \sqrt{y}$ A flat strip runs from the curve across to the line $x = 2$.
$y$ runs from 0 to $2^2 = 4$ The limits are now heights: the flat strips stack from $y = 0$ up to $y = 4$.
$A = \int_0^4 (2 - \sqrt{y}) \mathrm{d} y$ Add the flat strips: integrate the length with respect to $y$.
$A = [2y - (2/3) y^{3/2}]_0^4$ Write $\sqrt{y}$ as $y^{1/2}$, add 1 to the power, and divide by $3/2$.
$A = 8 - (2/3) \cdot 8 = 8 - 16/3 = 8/3$ Put in 4, then 0: $4^{3/2} = 8$, and at 0 every term is 0.
Check: both ways give $8/3$ Upright and flat strips cut up the same region, so the two areas must agree.
So you can choose which way the strips run. Pick the way that gives the easier integral.
Your turn: take the region between $x = y^2$, the $y$-axis and the lines $y = 0$ and $y = 2$. Which strips give you one easy integral?
Check yourself
For $f(x) \geq 0$ on $[a, b]$, the area bounded by $y = f(x)$, the x-axis, and the ordinates $x = a$, $x = b$ is \_\_\_\_\_\_.
$f(b) - f(a)$
$\int_a^b f(x) d x$
$(b - a) f(a)$
$\int_a^b f'(x) d x$
Check your answer
$f(b) - f(a)$ — $f(b) - f(a)$ tracks how much $f$ itself changed, not the region’s area. The area comes from integrating $f$, not from subtracting two of its values.
✓ $\int_a^b f(x) d x$ — (B) This is the direct statement of the area-under-a-curve formula: integrate $f$ from $a$ to $b$.
$(b - a) f(a)$ — A single rectangle using only $f(a)$ is a rough estimate, not the exact area. The exact formula integrates $f$ across the whole interval.
$\int_a^b f'(x) d x$ — The area formula integrates $f$ directly. Integrating $f'$ instead gives $f(b) - f(a)$ by the fundamental theorem, a different quantity, not the area under $f$.
A region is bounded by $x = g(y)$ with $g(y) \geq 0$, the y-axis, and $y = c$, $y = d$. Its area is \_\_\_\_\_\_.
$\int_a^b g(y) \mathrm{d} x$
$\int_c^d g(y) \mathrm{d} y$
$\int_c^d y \mathrm{d} x$
$g(d) - g(c)$
Check your answer
$\int_a^b g(y) \mathrm{d} x$ — The y-axis region’s bounds are $c$ and $d$, not $a$ and $b$, those belong to the x-axis version. Mixing the two versions’ bounds and variables does not describe either region correctly.
✓ $\int_c^d g(y) \mathrm{d} y$ — (B) The same idea works against the y-axis: each flat strip has length $g(y)$ and thickness $\mathrm{d} y$, so integrate $g(y)$ with respect to $y$ between the given $y$-bounds.
$\int_c^d y \mathrm{d} x$ — Each flat strip has length $g(y)$ and thickness $\mathrm{d} y$. Integrating $y$ against $x$ is the x-strip habit carried over, and it does not measure this region.
$g(d) - g(c)$ — $g(d) - g(c)$ tracks how much $g$ changed, not the area of the region against the y-axis. The area still comes from integrating.
Why does the y-axis area formula $\int_c^d g(y) d y$ use $d y$ rather than $d x$?
$g(y)$ is always negative, and $d y$ corrects the sign
$d y$ is just a stylistic alternative to $d x$, with no real difference
the y-axis version never uses strips at all
the strips are now horizontal, sliced along the y-axis
Check your answer
$g(y)$ is always negative, and $d y$ corrects the sign — $d y$ is a width, not a sign correction, it carries no information about whether $g(y)$ is positive or negative.
$d y$ is just a stylistic alternative to $d x$, with no real difference — The switch to $d y$ is not stylistic, it reflects that the strip is now horizontal, sliced along $y$, not vertical and sliced along $x$.
the y-axis version never uses strips at all — The y-axis version still builds the area from thin strips, they are simply horizontal now, running in the $y$ direction, rather than vertical.
✓ the strips are now horizontal, sliced along the y-axis — (D) Against the y-axis, the elementary strip lies on its side: its width is a tiny slice of $y$, not $x$, which is exactly why the differential switches to $d y$.
A new plot with no road. A curved footpath runs along the bottom, and the canal bank runs along the top.
Now change the plot. There is no road this time. Along the bottom runs a footpath, and the path curves too. The canal bank is still along the top. Mark the land into thin upright strips again.
Strips from the path up to the bank. Each is the bank’s height minus the path’s; eight of them add to about 14.69.
Each strip now starts on the path, not on the road. It ends on the bank. So its length is the height of the bank minus the height of the path.
strip at x
→
top minus bottom
→
× dx, add
from a to b
→
A = ∫ (f(x) − g(x)) dx
The rule between two curves: one strip is top minus bottom, f(x) − g(x). Add the strips from a to b.
That gives the rule. Let $f$ and $g$ be continuous on $[a, b]$, with $f(x) \geq g(x)$ at every $x$ there. So the curve $y = f(x)$ stays on top. The strip at $x$ runs from $g(x)$ up to $f(x)$. Its height is $f(x) - g(x)$. The AREA BETWEEN TWO CURVES is
$A = \int_a^b (f(x) - g(x)) \mathrm{d} x$.
Between two curves on 0 to 4, the strip at x equals 2 runs from 0.5 to 4.5. The band’s area is 44 over 3.
under the bank
under the path
52/3 − 8/3 = 44/3
Top minus bottom as two areas. Under the bank is 52/3, under the path 8/3, and the plot between them is 44/3.
Let us check it on the figure. The top curve is $y = 5 - x^2/8$, and the bottom one is $y = x^2/8$, from 0 to 4.
Worked example
The area between $y = 5 - x^2/8$ and $y = x^2/8$, from 0 to 4
$(5 - 2^2/8) - 2^2/8 = 4.5 - 0.5 = 4$ At $x = 2$ the strip runs from the bottom curve at 0.5 up to the top one at 4.5.
$(5 - x^2/8) - x^2/8 = 5 - x^2/4$ Subtract the bottom curve from the top one: the two $x^2/8$ terms make $x^2/4$.
$A = \int_0^4 (5 - x^2/4) \mathrm{d} x$ Add the strips from $x = 0$ to $x = 4$.
$A = [5x - x^3/12]_0^4$ Integrate term by term: $x^2/4$ gives $x^3/12$.
$A = 20 - 64/12 = 20 - 16/3$ Put in 4, then 0, and subtract: at 0 every term is 0.
*$A = 44/3 \approx 14.67$ square units* Write 20 as $60/3$ and subtract.
Check: $0 < 44/3 < 4 \times 5 = 20$ The area is positive and fits inside the rectangle 4 wide and 5 tall round the region.
The same band twice. Top minus bottom gives 44 over 3. Bottom minus top gives minus 44 over 3.
The order matters. Top minus bottom gives $44/3$. Bottom minus top gives $-44/3$, the same size with a minus sign. A minus or zero answer means the order is reversed, or a crossing inside the interval was not split. Check for crossings before you flip the sign.
Move both curves 4 down, so both dip below the axis. Every strip keeps its length, and the area stays 44/3.
The rule still holds if both curves dip below the axis. The strip’s height is still top minus bottom.
Check yourself
When $f(x) \geq g(x)$ on $[a, b]$, the area between the two curves is \_\_\_\_\_\_.
$\int_a^b (f(x) + g(x)) d x$
$\int_a^b (g(x) - f(x)) d x$
$\int_a^b f(x) g(x) d x$
$\int_a^b (f(x) - g(x)) d x$
Check your answer
$\int_a^b (f(x) + g(x)) d x$ — Adding $f$ and $g$ combines both curves’ heights, it does not isolate the strip of area BETWEEN them, which needs subtraction.
$\int_a^b (g(x) - f(x)) d x$ — $f$ is on top here, so it must come FIRST in the subtraction. Reversing the order makes the integral come out negative.
$\int_a^b f(x) g(x) d x$ — The region between two curves comes from SUBTRACTING the lower curve from the upper one, not multiplying them together.
✓ $\int_a^b (f(x) - g(x)) d x$ — (D) With $f$ above $g$, the area between them is the top curve minus the bottom curve, integrated across the interval.
Why must you identify which curve is on top BEFORE subtracting?
subtracting in the wrong order flips the sign
the bottom curve contributes nothing to the area
integration only works when applied to the top curve
only the top curve is continuous
Check your answer
✓ subtracting in the wrong order flips the sign — (A) The subtraction $f - g$ only gives a positive strip area when $f$ is genuinely the top curve; get the order backwards and every value flips sign.
the bottom curve contributes nothing to the area — The bottom curve’s height is subtracted, not ignored, it is exactly what carves the region OUT from under the top curve.
integration only works when applied to the top curve — Integration applies equally to both curves, the concern is purely about which one is subtracted from which, not about integration refusing to work on the bottom curve.
only the top curve is continuous — Continuity does not depend on which curve is drawn higher. It is enough for the integral to exist (Chapter 7), and it is not a separate condition here.
A student computes the area between $y = f(x)$ (below) and $y = g(x)$ (above) as $\int_a^b (f(x) - g(x)) d x$ and gets a negative number. What went wrong?
the bounds were entered in reverse order
the curves were subtracted in the wrong order
the integration was carried out with an arithmetic mistake
nothing is wrong; negative areas are normal here
Check your answer
the bounds were entered in reverse order — Nothing in the setup suggests the bounds were swapped, the subtraction itself has the curves in the wrong order, which directly causes the negative sign here.
✓ the curves were subtracted in the wrong order — (B) With $g$ above $f$, the top-minus-bottom subtraction is $g - f$. Writing it as $f - g$ reverses the order and flips the sign.
the integration was carried out with an arithmetic mistake — The negative sign is fully explained by the reversed subtraction order, there is no need to invoke a separate arithmetic slip on top of it.
nothing is wrong; negative areas are normal here — Area is positive by definition. A negative result here is the SIGNAL that the curves were subtracted in the wrong order, not an acceptable outcome to leave as is.
Two curves are given with no bounds $x = a$, $x = b$ stated. What must happen before $\int_a^b (f(x) - g(x)) d x$ can be applied?
the formula cannot be used at all without given bounds
the derivative of both curves must be set equal
the x-intercepts of $f$ alone must be found
the bounds come from the curves’ intersection points
Check your answer
the formula cannot be used at all without given bounds — The formula still works, it is simply missing its bounds, and those come from solving $f(x) = g(x)$, not from abandoning the formula.
the derivative of both curves must be set equal — Setting derivatives equal finds where the two curves have the same SLOPE, not where they physically meet. Meeting requires $f(x) = g(x)$.
the x-intercepts of $f$ alone must be found — The x-intercepts of $f$ tell you where $f$ crosses the x-axis, not where $f$ and $g$ meet each other. The bounds come from $f(x) = g(x)$, not from either curve’s own intercepts.
✓ the bounds come from the curves’ intersection points — (D) With no bounds given directly, the two curves’ own intersection points, found by solving $f(x) = g(x)$, become the bounds $a$ and $b$.
The path curves up and meets the bank twice. The land between them is one closed piece, pointed at both ends.
Some plots have no straight ends at all. The path curves up and meets the canal bank at two points. The land between them is one closed piece, pointed at both ends. It starts where the path meets the bank, and it stops where they meet again.
A line and a parabola cross at x equals 1 and x equals 4. The band between them has area 9 over 8.
So when no limits are given, the curves give them. The points where the two edges meet are the limits.
y equals x plus 4 meets y equals x squared over 2 at minus 2 and 4. The band between them is 18.
To find them, set the two heights equal. Solve $f(x) = g(x)$ for $x$. Each answer is a point where the curves meet. The smallest and the largest are the limits of the integral. The curves may also meet at a point between those two. Where $f(x) - g(x)$ changes sign there, the top and bottom curves swap, so split the integral at that point. Curves that only touch do not swap.
x²/4 + 1 = 5x/4
→
× 4
→
factor
→
read off
→
x = 1, x = 4
Set the heights equal, multiply by 4 and factor. The line and the parabola meet at x = 1 and x = 4.
Solving for the crossing
Let us find the limits for the figure. The line is $y = 5x/4$, and the parabola is $y = x^2/4 + 1$.
Worked example
Where $y = 5x/4$ meets $y = x^2/4 + 1$
$x^2/4 + 1 = 5x/4$ Set the two heights equal: where the curves meet, they have the same $y$.
$x^2 + 4 = 5x$ Multiply both sides by 4 to clear the fractions.
$x^2 - 5x + 4 = 0$ Take $5x$ from both sides, so that one side is 0.
$(x - 1)(x - 4) = 0$ Factor: $-1$ and $-4$ multiply to 4 and add to $-5$.
*$x = 1$ or $x = 4$* A product is 0 only when one of its brackets is 0.
Check: at $x = 4$, $5 \cdot 4/4 = 5$ and $4^2/4 + 1 = 5$ Both curves are at height 5 there, so they do meet.
Test x = 2: the line is at 2.5 and the parabola at 2, so the line is on top. The area is 9/8.
Now test a point between them. At $x = 2$ the line is at 2.5 and the parabola at 2. So the line is on top, and each strip is the line minus the parabola.
Worked example
The area between $y = 5x/4$ and $y = x^2/4 + 1$
$A = \int_1^4 (5x/4 - x^2/4 - 1) \mathrm{d} x$ Integrate the line minus the parabola, from one meeting point to the other.
$A = [5x^2/8 - x^3/12 - x]_1^4$ Integrate term by term.
At $x = 4$: $10 - 16/3 - 4 = 2/3$ Put in the top limit: $5 \cdot 16/8 = 10$ and $64/12 = 16/3$.
At $x = 1$: $5/8 - 1/12 - 1 = -11/24$ Put in the bottom limit, over the common denominator 24.
*$A = 2/3 + 11/24 = 27/24 = 9/8$ square units* Subtract the value at 1 from the value at 4.
Check: $0 < 9/8 < 3 \times 5 = 15$ The area is positive and fits inside the rectangle 3 wide and 5 tall round the region.
Left: guessed limits 0 to 3 give 3 over 8. Right: the crossings, 1 and 4, give 9 over 8.
Guessed limits give a wrong answer. With 0 and 3 you get $3/8$. You count a piece outside the region as negative, and you miss a piece inside it.
Your turn: where do $y = x$ and $y = x^3$ meet? Which points are the limits, and where do the top and bottom curves swap?
Check yourself
When two curves are given but no bounds $x = a$, $x = b$, where do the bounds come from?
the y-intercepts of both curves
the x-intercepts of $f$ alone
the curves’ own intersection points
the vertex of whichever curve is a parabola
Check your answer
the y-intercepts of both curves — The y-intercepts only show each curve’s own value at $x = 0$, they say nothing about where the two curves meet each other.
the x-intercepts of $f$ alone — The x-intercepts of $f$ show where $f$ crosses the x-axis, not where $f$ meets $g$. Solving $f(x) = g(x)$ is the actual bound-finding step.
✓ the curves’ own intersection points — (C) The natural bounds are exactly where the two curves cross each other, solve $f(x) = g(x)$ for $x$ to find them.
the vertex of whichever curve is a parabola — The vertex belongs to one curve alone. Where two curves cross depends on both of them together, through $f(x) = g(x)$.
For the curves $y = 2x$ and $y = x^2$, solving for their intersection points gives which bounds?
$x = 0$ and $x = 1$
$x = 0$ and $x = 2$
$x = 1$ and $x = 2$
$x = -2$ and $x = 0$
Check your answer
$x = 0$ and $x = 1$ — Factoring $x \cdot (x - 2) = 0$ gives roots $0$ and $2$, not $1$. Re-check the factoring step.
$x = 1$ and $x = 2$ — Factoring gives two roots, $0$ and $2$, dropping the zero root leaves only one of the two correct bounds.
$x = -2$ and $x = 0$ — The factoring $x \cdot (x - 2) = 0$ gives $x = 2$, not $x = -2$. The sign of the nonzero root was flipped.
Why do the solutions of $f(x) = g(x)$ work as the bounds for the area between the curves?
exactly where the two curves meet each other
they are where each curve individually crosses the x-axis
they are where $f$ or $g$ is undefined
they are the average of $f$ and $g$’s own separate roots
Check your answer
✓ exactly where the two curves meet each other — (A) Between two intersection points, one curve stays above the other; outside them, the region between the curves has not yet started or has already ended.
they are where each curve individually crosses the x-axis — Crossing the x-axis solves $f(x) = 0$. Meeting the OTHER curve solves $f(x) = g(x)$, a different equation, giving a different set of points.
they are where $f$ or $g$ is undefined — At an intersection point both $f$ and $g$ are defined and equal, nothing about $f(x) = g(x)$ points to where either function fails to exist.
they are the average of $f$ and $g$’s own separate roots — There is no averaging step here. The intersection points come directly from solving $f(x) = g(x)$, using both curves together, not from averaging each curve’s own roots.
Below the axis every strip has a negative height. Four strips under y = x²/2 − 2 from 0 to 2 add to about −2.69.
An area is never negative. A definite integral can be. Where $f(x) < 0$, each strip has a negative height. So the integral counts that piece as negative. This is the same idea as Chapter 7’s tank with the tap reversed: water going out counts against water coming in.
Wholly below the axis from 0 to 2, the integral is −8/3. Its absolute value, 8/3, is the area.
Say the curve is below the axis on all of $[a, b]$. Then the integral is negative. The rule is to take its absolute value. The area is $\left|\int_a^b f(x) \mathrm{d} x\right|$.
Split where the curve crosses the axis, at x equals 2. The pieces are minus 8 over 3 and 16 over 3. Area: 8.
Now say the curve crosses the axis at a point $c$ inside $(a, b)$. Then one integral from $a$ to $b$ mixes a plus piece and a minus piece. Split it at the crossing, and add the sizes.
Find the crossing first. x²/2 − 2 = 0 gives x = −2 or 2, and only 2 lies inside [0, 4].
Let us split the curve in the figure, $y = x^2/2 - 2$ on $[0, 4]$.
Worked example
The area between $y = x^2/2 - 2$ and the $x$-axis on $[0, 4]$
$x^2/2 - 2 = 0$ Set the height equal to 0 to find where the curve crosses the axis.
$x^2 = 4$ Add 2 to both sides, then multiply both sides by 2.
$x = 2$ Take the square root: of 2 and $-2$, only 2 lies inside $[0, 4]$.
$\int_0^2 (x^2/2 - 2) \mathrm{d} x = [x^3/6 - 2x]_0^2 = 4/3 - 4 = -8/3$ From 0 to 2 the curve is below the axis, so this integral is negative.
$\int_2^4 (x^2/2 - 2) \mathrm{d} x = [x^3/6 - 2x]_2^4 = 8/3 - (-8/3) = 16/3$ From 2 to 4 the curve is above the axis: put in 4, then 2, and subtract.
$\left|-8/3\right| + \left|16/3\right| = 8/3 + 16/3$ Take the size of each piece: the absolute value drops the minus sign.
*$A = 24/3 = 8$ square units* Add the two sizes.
Check: $0 < 8 < 4 \times 8 = 32$ The area is positive and fits inside the rectangle 4 wide, from $y = -2$ up to $y = 6$.
One curve, two counts. Left, the integral: minus 8 over 3 plus 16 over 3, which is 8 over 3. Right, the area: 8.
One integral straight across gives $-8/3 + 16/3 = 8/3$. That is not the area.
y equals x and y equals x cubed over 4 cross at minus 2, 0 and 2. Each lobe is 1: area 2, not 0.
Two curves work the same way. Split wherever they cross each other.
Your turn: find the area between $y = x^2 - 1$ and the $x$-axis, from $x = 0$ to $x = 2$. Where does the curve cross the axis? What does one integral straight across give?
Check yourself
By definition, is area allowed to be negative?
no: area is always positive; the integral can still be negative
yes: area can be negative whenever the curve dips below the x-axis
yes, but only for regions bounded by a circle or a parabola
no: but only because the integral is always positive too
Check your answer
✓ no: area is always positive; the integral can still be negative — (A) Area is positive by definition. The definite integral is signed, and comes out negative wherever the curve (or $f - g$) is negative, the two are not the same quantity.
yes: area can be negative whenever the curve dips below the x-axis — The integral can go negative below the axis, but AREA never does, area is positive by definition, whatever the integral’s raw sign happens to be.
yes, but only for regions bounded by a circle or a parabola — Area is positive by definition for every shape, circles, parabolas, or anything else. There is no shape-based exception.
no: but only because the integral is always positive too — The integral is NOT always positive, it goes negative wherever the curve does. Area’s positivity is a separate, unconditional definition, not a consequence of the integral always being positive.
If $f$ changes sign at some $c$ in $(a, b)$, what must you do to find the true area?
split at $c$, take the absolute value of each piece, then add them
integrate straight across and take the absolute value of the single final number
integrate only the positive portion and ignore the negative portion
ignore the sign and report the raw integral
Check your answer
✓ split at $c$, take the absolute value of each piece, then add them — (A) Splitting at the sign change isolates a positive piece and a negative piece; taking the absolute value of each before adding recovers the true, all-positive area.
integrate straight across and take the absolute value of the single final number — By the time you reach the final number, the positive and negative pieces have already cancelled inside the integral. Taking the absolute value of that already-cancelled result does not recover the true area.
integrate only the positive portion and ignore the negative portion — Both pieces of the region count toward the total area. Dropping the negative-valued piece entirely leaves out part of the true area, rather than correcting its sign.
ignore the sign and report the raw integral — The raw integral already has the negative piece cancelling the positive piece , reporting it as is skips the correction entirely.
For $f(x) = x$ on $[-2, 3]$, where must the integral be split to correctly compute the area?
at $x = 3$, the right endpoint
at $x = 0$
at $x = -2$, the left endpoint
at $x = 0.5$, the midpoint of $[-2, 3]$
Check your answer
at $x = 3$, the right endpoint — $f(3) = 3$, already positive, the sign changed before this endpoint was reached. The flip happens at $x = 0$.
✓ at $x = 0$ — (B) $f(x) = x$ changes sign exactly at $x = 0$, the only point in $(-2, 3)$ where the curve crosses the x-axis.
at $x = -2$, the left endpoint — $f(-2) = -2$, still negative, the sign has not changed yet at the left endpoint. The flip happens at $x = 0$.
at $x = 0.5$, the midpoint of $[-2, 3]$ — The midpoint of the interval is not where the function’s sign changes, that depends on where $f(x) = 0$, which for $f(x) = x$ is $x = 0$, not the interval’s midpoint.
After splitting $\int_{-2}^3 x d x$ at $x = 0$, how do you correctly combine the two pieces to get the true area?
subtract the second piece from the first
take the absolute value of each piece separately, then add them
add the two pieces first, then take the absolute value of that sum
add the two pieces directly, with no absolute value at all
Check your answer
subtract the second piece from the first — Both split pieces are parts of the SAME region’s area, they are added together, once each is made positive, never subtracted from one another.
✓ take the absolute value of each piece separately, then add them — (B) Each piece is made positive on its own before adding, so a negative piece and a positive piece both contribute their full size to the total, rather than cancelling.
add the two pieces first, then take the absolute value of that sum — Taking the absolute value of the SUM still lets the pieces cancel each other first. Each piece needs the absolute value applied to IT individually, before adding.
add the two pieces directly, with no absolute value at all — Adding the signed pieces directly still lets the negative piece cancel part of the positive one, the split alone does not fix that; each piece needs its OWN absolute value first.
A round flower bed in a park. Class 10 gave its area as π times the radius squared, but never showed why.
In Class 10 you found the area of many circles. You used one formula, pi times the square of the radius. You were told that formula, but nobody showed you why it is true. The strips can show you.
The circle of radius 2 cut into upright strips. Above the x-axis each strip reaches the top half, y = √(4 − x²).
A circle is a region with a curved edge, so cut it into upright strips. Take the circle $x^2 + y^2 = a^2$, of radius $a$ about the origin. Take $x^2$ from both sides:
$y^2 = a^2 - x^2$.
The top half has $y \geq 0$, so take the positive square root:
$y = \sqrt{a^2 - x^2}$.
A circle of radius 2, cut into four quarters by the axes. Each quarter is pi, so the whole circle is 4 pi.
The two axes cut the circle into four equal quarters. So find the quarter in the first quadrant, and multiply by 4.
$A = 4 \int_0^a \sqrt{a^2 - x^2} \mathrm{d} x$.
√(a² − x²)
→
integrate
Chapter 7
→
(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C
Chapter 7’s integral of √(a² − x²). Differentiate the answer and you get the integrand back.
The first quarter, under y = √(4 − x²) from 0 to 2. Its area is π, which is π × 2² ÷ 4.
Why it works
Find the first quarter, $\int_0^a \sqrt{a^2 - x^2} \mathrm{d} x$
$(a/2) \sqrt{a^2 - a^2} = 0$ Put $x = a$ into the first term: the root is $\sqrt{0} = 0$.
$(a^2/2) \sin^{-1}(a/a) = (a^2/2) \sin^{-1}(1)$ Put $x = a$ into the second term: $a/a = 1$.
$(a^2/2) \cdot \pi/2 = (\pi a^2)/4$ The angle whose sine is 1 is $\pi/2$, so $\sin^{-1}(1) = \pi/2$.
$(0/2) \sqrt{a^2} + (a^2/2) \sin^{-1}(0) = 0$ Put $x = 0$ in: the first term has a factor 0, and $\sin^{-1}(0) = 0$.
*$\int_0^a \sqrt{a^2 - x^2} \mathrm{d} x = (\pi a^2)/4 - 0 = (\pi a^2)/4$* Subtract the value at 0 from the value at $a$.
Four quarters make the whole circle:
$A = 4 \cdot (\pi a^2)/4 = \pi a^2$.
Four quarters of π make the whole disc, 4π ≈ 12.57. That is π × 2², the Class 10 formula.
That is the Class 10 formula, with $a$ for the radius. Let us check it with $a = 2$. The quarter is $\pi$, and four quarters make $4 \pi$. That is $\pi \cdot 2^2$.
Your turn: use the same steps for the circle $x^2 + y^2 = 9$. What is one quarter, and what is the whole area?
A SCHOLAR INDIA REMEMBERS
How big is the circle of radius $2$, before any formula? Bound it between two squares. The square round it is $4$ by $4$, so its area is $16$. The square inside it has its corners on the circle, and its diagonals are $4$ long. Its area is $4 \times 4 / 2 = 8$. So the circle is more than $8$ and less than $16$. The strips give $4 \pi$, about $12.57$, between the bounds, as it must be.
Aryabhata · the hoopoe
Check yourself
For the circle $x^2 + y^2 = a^2$, what is the equation of its upper half?
$y = \sqrt{a^2 - x^2}$
$y = \sqrt{a^2 + x^2}$
$y = \sqrt{a - x^2}$
$y = a - x$
Check your answer
✓ $y = \sqrt{a^2 - x^2}$ — (A) Solving $x^2 + y^2 = a^2$ for $y$ and keeping the positive root gives the upper half, $y = \sqrt{a^2 - x^2}$.
$y = \sqrt{a^2 + x^2}$ — Solving $x^2 + y^2 = a^2$ for $y$ gives $y^2 = a^2 - x^2$, a subtraction, not $a^2 + x^2$. The sign inside the root was flipped.
$y = \sqrt{a - x^2}$ — The circle’s equation has $a^2$, not $a$, so solving for $y$ keeps the square: $y = \sqrt{a^2 - x^2}$.
$y = a - x$ — The circle’s upper half is a curve, not a straight line, $y = a - x$ describes a line, not any part of $x^2 + y^2 = a^2$.
Why does the full circle’s area come out as $4 \int_0^a \sqrt{a^2 - x^2} d x$ rather than the integral alone?
the circle has 4 sides, one per axis crossing
every circle-area computation always needs a factor of 4, regardless of the region computed
the integral covers one quarter; the other three quarters match it by symmetry
$\pi$ is being approximated using 4 rectangles
Check your answer
the circle has 4 sides, one per axis crossing — A circle is a smooth curve with no sides. The factor of 4 comes from its four-fold symmetry, not from anything resembling sides.
every circle-area computation always needs a factor of 4, regardless of the region computed — The factor of 4 is not an arbitrary rule, it comes directly from this specific integral covering exactly one of the circle’s four symmetric quarters.
✓ the integral covers one quarter; the other three quarters match it by symmetry — (C) $\int_0^a \sqrt{a^2 - x^2} d x$ is the area of just one quarter-circle; the circle’s symmetry means the other three quarters match it exactly, so multiplying by 4 gives the whole circle.
$\pi$ is being approximated using 4 rectangles — No approximation or rectangles are involved. The factor of 4 comes from an exact symmetry argument: one quarter’s area, times four identical quarters.
For $x^2 + y^2 = 9$, what integral (before evaluating) gives the full circle’s area, by the symmetry method used in this chapter?
$4 \int_0^3 \sqrt{3 - x^2} d x$
$\int_0^3 \sqrt{9 - x^2} d x$
$4 \int_0^9 \sqrt{9 - x^2} d x$
$4 \int_0^3 \sqrt{9 - x^2} d x$
Check your answer
$4 \int_0^3 \sqrt{3 - x^2} d x$ — Inside the square root the formula needs $a^2 - x^2$, which here is $9 - x^2$, using $a$ alone, $3 - x^2$, drops the square.
$\int_0^3 \sqrt{9 - x^2} d x$ — Without the factor of 4, this integral is only ONE quarter-circle’s area, the whole circle needs all four identical quarters counted.
$4 \int_0^9 \sqrt{9 - x^2} d x$ — The upper bound of integration is $a = 3$, the radius, not $a^2 = 9$, the right-hand side of the circle’s own equation.
✓ $4 \int_0^3 \sqrt{9 - x^2} d x$ — (D) Here $a^2 = 9$, so $a = 3$; the method integrates $\sqrt{9 - x^2}$ from $0$ to $3$ and multiplies by 4 for the four symmetric quarters.
For the general circle $x^2 + y^2 = a^2$, evaluating $4 \int_0^a \sqrt{a^2 - x^2} d x$ always simplifies to \_\_\_\_\_\_.
$\pi a^2$
$(4/3) \pi a^3$
$2 \pi a$
$\pi a$
Check your answer
✓ $\pi a^2$ — (A) This is the familiar circle-area formula, now derived from the integral rather than assumed.
$(4/3) \pi a^3$ — $(4/3) \pi a^3$ is a sphere’s volume, a completely different three-dimensional formula. This integral computes a flat area, $\pi a^2$.
$2 \pi a$ — $2 \pi a$ is the circle’s circumference, a length, not its area. The area, derived from this integral, is $\pi a^2$.
$\pi a$ — An area formula needs a squared length, $a^2$, not $a$ alone. $\pi a$ has the wrong units to be an area.
▲CONCEPT
An oval cricket ground from above. Its edge is close to an ellipse: a circle stretched one way.
The circle of radius 2 with every height halved. The result is the ellipse x²/4 + y² = 1, 2 across and 1 up.
An ELLIPSE is a circle stretched one way. Start with the circle of radius $a$, and multiply every height by $b/a$, with $a > 0$ and $b > 0$. You get the ellipse $x^2/a^2 + y^2/b^2 = 1$. It reaches $a$ along the $x$-axis and $b$ along the $y$-axis.
the circle
the ellipse
At x = 1 the circle’s strip is √3 ≈ 1.73 tall and the ellipse’s is half that, 0.87. Every strip is b/a times.
Its top half is $y = (b/a) \sqrt{a^2 - x^2}$. That is the circle’s top half, with every height times $b/a$. So each upright strip is $b/a$ times the circle’s strip at the same $x$.
The ellipse’s first quarter is half the circle’s quarter: π/2. Four of them make 2π.
The axes cut the ellipse into four equal quarters too. So its area is 4 times the first quarter.
Each strip is $b/a$ times the circle’s strip. A constant factor comes out in front of an integral, so the first quarter is $b/a$ times the circle’s quarter:
The circle of radius 2 is 4π. The ellipse with a = 2, b = 1 is π × 2 × 1 = 2π, half.
When $b = a$, the ellipse is a circle, and $\pi a b$ becomes $\pi a^2$. Check it with $a = 2$ and $b = 1$. The area is $\pi \cdot 2 \cdot 1 = 2 \pi$, half the circle of radius 2.
Your turn: find the area of the ellipse $x^2/16 + y^2/9 = 1$. What are $a$ and $b$?
A SCHOLAR INDIA REMEMBERS
Does the circle’s method work for the next case too? Try the ellipse with $a = 2$ and $b = 1$. At $x = 1$ the circle’s strip is $\sqrt{3}$ tall, about $1.73$. The ellipse’s strip there is half of that, about $0.87$. Every strip is halved, so the area is halved too, and $4 \pi$ becomes $2 \pi$. The method carried over whole. The new shape cost one extra factor, $b/a$.
Aryabhata · the hoopoe
▲CONCEPT
latus rectum x = 1
The parabola y² = 4x opens to the right from the origin. Its latus rectum, the line x = 1, closes off a region.
Take the parabola $y^2 = 4 a x$, with $a > 0$. It opens to the right from the origin. Its LATUS RECTUM is the line $x = a$. That line and the parabola close off a region.
Solve for y. The top arc is y = 2√x and the bottom arc y = −2√x. The x-axis is a mirror between them.
The region has a top arc and a bottom arc. Solve $y^2 = 4 a x$ for $y$ by taking the square root. The positive root is the top arc:
$y = \sqrt{4 a x} = 2 \sqrt{a x}$.
The negative root is the bottom arc:
$y = -2 \sqrt{a x}$.
The mirror line is the $x$-axis, since $y$ and $-y$ give the same $y^2$. So the region is twice its top half.
y² = 4x
y² = 6x
Cut off by its latus rectum, y² = 4x gives 8/3 and y² = 6x gives 6. Both are (8/3)a².
Why it works
The area to the latus rectum of $y^2 = 4 a x$
$A = 2 \int_0^a 2 \sqrt{a x} \mathrm{d} x$ Double the area under the top arc $y = 2 \sqrt{a x}$, from $x = 0$ to the latus rectum.
$A = 4 \sqrt{a} \int_0^a \sqrt{x} \mathrm{d} x$ Take the constants out of the integral: $2 \cdot 2 = 4$ and $\sqrt{a x} = \sqrt{a} \sqrt{x}$.
$\int_0^a \sqrt{x} \mathrm{d} x = [(2/3) x^{3/2}]_0^a = (2/3) a^{3/2}$ Write $\sqrt{x}$ as $x^{1/2}$, add 1 to the power, and divide by $3/2$.
$A = 4 \sqrt{a} \cdot (2/3) a^{3/2}$ Put this integral back into $A$.
Left: the parabola y squared equals 4x, cut off at x equals 1, area 8 over 3. Right: the top half, 4 over 3.
Check it with $a = 1$. The parabola is $y^2 = 4 x$, and the latus rectum is $x = 1$. The top arc is $y = 2 \sqrt{x}$, so the top half is:
$\int_0^1 2 \sqrt{x} \mathrm{d} x = [(4/3) x^{3/2}]_0^1 = 4/3$.
Double it for the whole region:
$2 \cdot 4/3 = 8/3$.
That matches the formula $(8/3) a^2$ at $a = 1$.
y² = 4x
double the top half
(y − 1)² = 4x
Doubling the arc above the x-axis takes the x-axis as the mirror: 14/3. The real mirror is y = 1, and the area is 8/3.
Before you double a half, check that the mirror is real. Put $-y$ in place of $y$ in the equation. If nothing changes, the two halves match.
Your turn: find the area cut off by the latus rectum of $y^2 = 8 x$. What is $a$?
Check yourself
For the parabola $y^2 = 4 a x$, what is the equation of its upper arc?
$y = 4 \sqrt{a x}$
$y = \sqrt{2 a x}$
$y = 2 \sqrt{a x}$
$y = 2 \sqrt{a + x}$
Check your answer
$y = 4 \sqrt{a x}$ — $\sqrt{4 a x} = \sqrt{4} \sqrt{a x} = 2 \sqrt{a x}$, the 4 inside the root becomes a 2 outside it, not a 4.
$y = \sqrt{2 a x}$ — The parabola’s own equation has $4 a x$ under the root, not $2 a x$. Pulling the square root of 4 out front, correctly, gives $2 \sqrt{a x}$.
✓ $y = 2 \sqrt{a x}$ — (C) Solving $y^2 = 4 a x$ for $y$ and keeping the positive root gives $y = \sqrt{4 a x} = 2 \sqrt{a x}$.
$y = 2 \sqrt{a + x}$ — The parabola’s equation multiplies $a$ and $x$ together, $4 a x$, solving for $y$ keeps that multiplication inside the root, not an addition.
Why is the parabola’s area (bounded by $y^2 = 4 a x$ and the latus rectum $x = a$) computed as 2 times an integral, rather than 4 times as with the circle?
integrating $\sqrt{x}$ requires doubling the result by definition
parabolas always use a factor of 2 as a fixed rule, unrelated to symmetry
symmetric about the x-axis only, not four-fold like a circle
the latus rectum only covers half the region, regardless of symmetry
Check your answer
integrating $\sqrt{x}$ requires doubling the result by definition — There is no general rule that integrating a square root always needs doubling. The factor of 2 here comes specifically from this region’s top-bottom symmetry.
parabolas always use a factor of 2 as a fixed rule, unrelated to symmetry — The factor of 2 is not a fixed convention, it comes directly from this specific region’s symmetry, top half matching bottom half across the x-axis.
✓ symmetric about the x-axis only, not four-fold like a circle — (C) A circle is symmetric about both axes, giving four matching quarters. This parabola region only mirrors top to bottom across the x-axis, giving just two matching halves.
the latus rectum only covers half the region, regardless of symmetry — The latus rectum is one edge of the region, not a measure of how much area it contains. The factor of 2 comes from the region’s own top-bottom symmetry, not from the latus rectum.
For $y^2 = 8 x$ (so $4 a = 8$, $a = 2$), the latus rectum is at which value of $x$?
$x = 4$
$x = 8$
$x = 2$
$x = 1$
Check your answer
$x = 4$ — Solving $4 a = 8$ for $a$ means dividing by 4, giving $a = 2$, dividing by 2 instead gives the wrong value, 4.
$x = 8$ — $8$ is the value of $4 a$, not $a$ itself. Dividing by 4 first gives $a = 2$, so the latus rectum is $x = 2$.
✓ $x = 2$ — (C) $4 a = 8$ gives $a = 2$, and the latus rectum is always the line $x = a$.
$x = 1$ — Solving $4 a = 8$ for $a$ means dividing by 4, giving $a = 2$, dividing by 8 instead gives the wrong value, 1.
For $y^2 = 4 a x$ with latus rectum $x = a$, the area bounded by the parabola and the latus rectum is \_\_\_\_\_\_.
$(8/3) a^2$
$(2/3) a^2$
$(4/3) a^2$
$\pi a^2$
Check your answer
✓ $(8/3) a^2$ — (A) This is the direct statement of the parabola-area formula, derived from integrating the upper arc and doubling for the top-bottom symmetry.
$(2/3) a^2$ — $(2/3) a^2$ is missing more than one factor of 2 from the actual derivation. The full, correctly derived area is $(8/3) a^2$.
$(4/3) a^2$ — $(4/3) a^2$ is only HALF the region, the top-bottom symmetry factor of 2 was dropped. The full area is $(8/3) a^2$.
$\pi a^2$ — $\pi a^2$ is the circle’s area formula. This region is bounded by a parabola, not a circle, and its area is $(8/3) a^2$.
The working for the area between y = x and the x-axis, from −1 to 1. Which step is wrong? Keep your answer.
Here is a working for the area between $y = x$ and the $x$-axis, from $x = -1$ to $x = 1$. Which step is wrong?
The area under a curve is its integral, so $A = \int_{-1}^1 x \mathrm{d} x$.
$\int_{-1}^1 x \mathrm{d} x = [x^2/2]_{-1}^1 = 1/2 - 1/2 = 0$.
So the area is 0.
y equals x from minus 1 to 1. Two triangles of 1 over 2 each make an area of 1, but the integral is 0.
Keep your answer, then look at the figure. The line makes two triangles with the axis. Each has base 1 and height 1, so each has area $1/2$. The one on $[-1, 0]$ is below the axis, and the integral counts it as $-1/2$. The one on $[0, 1]$ counts as $1/2$. The two cancel to 0.
Split at x = 0. The pieces are −1/2 and 1/2. Their sizes add to 1, the area; straight across, the integral is 0.
The first step is wrong. The area is the integral only where $f(x) \geq 0$. Here $x < 0$ on the left half. Split at the crossing $x = 0$, and add the sizes: $1/2 + 1/2 = 1$.
First try
I computed $\int_{-1}^1 x \mathrm{d} x = 0$ and wrote the area as $0$ square units.
Second look
Check the sign of the curve across the whole interval before trusting a raw integral as the area, exactly as the worked fix above does.
The true area is still not $0$: see the split just above.
Checking a zero integral
An area of 0 would mean there is no region at all.
Check yourself
Integrating straight across, so that the pieces cancel, breaks which general rule?
always double the final answer to account for symmetry
always integrate the derivative first, then the original function
split at every sign change and take the absolute value of each piece
always evaluate the integrand at the interval’s midpoint before integrating
Check your answer
always double the final answer to account for symmetry — Doubling is a rule specific to certain symmetric regions, like the circle and parabola area formulas, not a general fix for a sign-cancelling integral.
always integrate the derivative first, then the original function — There is no derivative-then-integrate step involved anywhere in this trap. The actual missing step is splitting at the sign change and taking absolute values.
✓ split at every sign change and take the absolute value of each piece — (C) The trap is exactly what happens when the split-and-absolute-value rule is skipped, the positive and negative pieces are left to cancel inside one uncorrected integral.
always evaluate the integrand at the interval’s midpoint before integrating — Evaluating at a midpoint has no bearing on this trap. The fix is splitting at the actual sign-change point and taking absolute values, not evaluating anywhere in particular.
The region under y = x² from 0 to 2 has area 8/3. It sits inside the 2 by 4 rectangle, area 8.
Worked example
Find the area bounded by $y = x^2$ and the ordinates $x = 0$, $x = 2$
$y = x^2$, x-axis, $x = 0$ to $x = 2$ $x^2 \geq 0$ on $[0, 2]$, so the curve stays on or above the axis. The integral is the area.
$A = \int_0^2 x^2 \mathrm{d} x$ The area under one curve is the integral of its height.
$A = [x^3/3]_0^2$ An integral of $x^2$ is $x^3/3$.
$A = (8/3) - 0 = 8/3$ Subtract the value at 0 from the value at 2.
*$A = 8/3$ square units* The area is one positive number.
Check: $0 < 8/3 < 2 \times 4 = 8$ The answer is positive. It fits inside the rectangle 2 wide and 4 tall round the region.
Worked example
Find the work done on a gas compressed from $10$ L to $4$ L against a constant external pressure of $2$ atm
$w = -p_{\text{ex}} (V_f - V_i)$, with $p_{\text{ex}} = 2$ atm, $V_i = 10$ L and $V_f = 4$ L Work is a signed integral, not an outright area; only its SIZE will equal a rectangle’s area here. Chemistry 11, Chapter 5, Thermodynamics, gives the work done on a gas in one irreversible step, compressed or expanded against a SINGLE constant external pressure $p_{\text{ex}}$.
On a pressure-volume graph, $p_{\text{ex}} = 2$ is a flat line, so the region under it from $V = 4$ to $V = 10$ is a rectangle A constant curve has a constant height, the simplest shape the area-under-a-curve idea builds from.
$A = \int_4^{10} 2 \mathrm{d} V = 2 \times 6 = 12$ An area runs from the smaller volume to the larger, so it is $p_{\text{ex}} \left|V_f - V_i\right|$; a constant integrand needs only Chapter 7’s constant rule.
*$w = -2 (4 - 10) = 12$ L atm* The work is the signed integral $-\int_{V_i}^{V_f} p_{\text{ex}} \mathrm{d} V$, not an area. Only its size equals the rectangle’s area.
Check: $-\int_{10}^4 2 \mathrm{d} V = \int_4^{10} 2 \mathrm{d} V = 12$ This is the rectangle’s area. Swapping the limits undoes the minus sign. The number is positive, matching work done ON the gas during compression.
Check yourself
Using the same method as the $y = x^2$ worked example, what is the area bounded by $y = x^3$, the x-axis, and $x = 0$, $x = 1$?
$4$ square units
$1$ square unit
$1/3$ square units
$1/4$ square units
Check your answer
$4$ square units — $4$ is the correct answer’s fraction flipped upside down. The power rule divides by the new exponent; it does not multiply by it.
$1$ square unit — $[x^4]_0^1 = 1$ is the raised power alone, the power rule for integration still needs dividing by the new exponent, giving $1/4$, not $1$.
$1/3$ square units — Integrating $x^3$ raises the power to $x^4$ and divides by the NEW exponent, 4, dividing by the original exponent, 3, gives the wrong antiderivative.
✓ $1/4$ square units — (D) $\int_0^1 x^3 d x = [x^4/4]_0^1 = 1/4$ square units.
y equals x lies above y equals x squared from 0 to 1. The strip at 0.6 is 0.24 long. Area: 1 over 6.
At x = 1/2 the line is at 1/2 and the parabola at 1/4. The region, 1/6, fits well inside the 1 by 1 square.
Worked example
Find the area between $y = x$ and $y = x^2$
$y = x$ and $y = x^2$ No limits are given. So first find where the two curves meet.
$x = x^2 \Rightarrow x \cdot (x - 1) = 0 \Rightarrow x = 0, 1$ Set the heights equal and factor. The curves meet at $x = 0$ and $x = 1$.
On $(0, 1)$, $x > x^2$ Test one point between the limits. At $x = 1/2$, $1/2 > 1/4$, so the line is on top.
$A = \int_0^1 (x - x^2) \mathrm{d} x$ Top minus bottom, from one meeting point to the other.
$A = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3$ Work out the integral at 1. At 0 it is 0.
*$A = 1/6$ square units* Subtract the two fractions.
Check: $0 < 1/6 < 1 \times 1 = 1$ The answer is positive. It fits inside the square 1 wide and 1 tall round the region.
Check yourself
For the area between $y = x$ and $y = x^2$, the computed area is \_\_\_\_\_\_.
$1/3$ square units
$5/6$ square units
$1/6$ square units
$1/2$ square units
Check your answer
$1/3$ square units — $1/3$ is only the SECOND term of the antiderivative. The first term, $x^2/2$ at $x = 1$, still has to be included, giving $1/2 - 1/3 = 1/6$.
$5/6$ square units — $5/6 = 1/2 + 1/3$ adds the two terms instead of subtracting the second from the first. The correct combination is $1/2 - 1/3 = 1/6$.
✓ $1/6$ square units — (C) $\int_0^1 (x - x^2) d x = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6$ square units.
$1/2$ square units — $1/2$ is only the FIRST term of the antiderivative. The second term, $x^3/3$ at $x = 1$, still has to be subtracted, giving $1/2 - 1/3 = 1/6$.
For the area between $y = 3x$ and $y = x^2$, solving for the intersection points gives which bounds?
$x = -3$ and $x = 0$
$x = 0$ and $x = 1$
$x = 1$ and $x = 3$
$x = 0$ and $x = 3$
Check your answer
$x = -3$ and $x = 0$ — Factoring $x \cdot (x - 3) = 0$ gives roots $0$ and $3$, not $-3$. The sign of the nonzero root was flipped.
$x = 0$ and $x = 1$ — $x = 0, 1$ solves $x = x^2$, a different pair of curves. Solving $x^2 = 3x$ for THIS pair gives $x = 0, 3$.
$x = 1$ and $x = 3$ — Factoring gives two roots, $x = 0$ and $x = 3$, dropping the zero root leaves only one of the two correct bounds.
For $y = 3x$ and $y = x^2$ on $[0, 3]$, with $3x$ above $x^2$, what is the area between them?
$-9/2$ square units
$9$ square units
$9/2$ square units
$27/2$ square units
Check your answer
$-9/2$ square units — $-9/2$ comes from subtracting in the reverse order, $x^2 - 3x$. With $3x$ on top, the subtraction must be $3x - x^2$.
$9$ square units — $9$ is only the SECOND term of the antiderivative, evaluated at $x = 3$. The first term, $(3x^2)/2$ at $x = 3$, still has to be included, giving $27/2 - 9 = 9/2$.
✓ $9/2$ square units — (C) $\int_0^3 (3x - x^2) d x = [(3x^2)/2 - x^3/3]_0^3 = 27/2 - 9 = 9/2$ square units.
$27/2$ square units — $27/2$ is only the FIRST term of the antiderivative, evaluated at $x = 3$. The second term, $x^3/3$ at $x = 3$, still has to be subtracted, giving $27/2 - 9 = 9/2$.
For radius 2: the algebraic term rises to 1 and falls to 0 at x equals 2. The inverse-sine term reaches pi, the whole quarter.
The circle of radius 2 has area 4π ≈ 12.57. It fits inside the 4 by 4 square round it, area 16.
Worked example
Find the area of the circle $x^2 + y^2 = 4$
$x^2 + y^2 = 4$, so $a = 2$ Read the radius off the equation. Here $a^2 = 4$, so $a = 2$.
$A = 4 \int_0^2 \sqrt{4 - x^2} \mathrm{d} x$ The axes cut the circle into four equal quarters. Find the first quarter and multiply by 4.
$\int \sqrt{a^2 - x^2} \mathrm{d} x = (x/2) \sqrt{a^2 - x^2} + (a^2/2) \sin^{-1}(x/a)$ This integral comes from Chapter 7.
$\int_0^2 \sqrt{4 - x^2} \mathrm{d} x = \pi$ At $x = 2$ the first term is 0 and the second is $2 \cdot \pi/2 = \pi$. At $x = 0$ both terms are 0.
*$A = 4 \pi$ square units* Four quarters make $4 \pi$. That is $\pi a^2$ for $a = 2$.
Check: $0 < 4 \pi \approx 12.57 < 4 \times 4 = 16$ The answer is positive. It fits inside the square 4 wide and 4 tall round the circle.
Check yourself
For $x^2 + y^2 = 4$, the total circle area computed in the worked example is \_\_\_\_\_\_.
$4 \pi$
$16 \pi$
$2 \pi$
$8 \pi$
Check your answer
✓ $4 \pi$ — (A) With $a = 2$, $\pi a^2 = \pi \cdot 4 = 4 \pi$, matching the worked example’s evaluated integral.
$16 \pi$ — The $4$ in the circle’s equation is $a^2$, not $a$, the radius is $a = 2$. Using $4$ as the radius gives $\pi \cdot 4^2 = 16 \pi$, the wrong quantity squared.
$2 \pi$ — $2 \pi$ uses $a = 2$ instead of $a^2 = 4$. The formula needs $a$ squared: $\pi \cdot 4 = 4 \pi$.
$8 \pi$ — The four-fold symmetry factor is already fully accounted for inside $\pi a^2$. Doubling the result applies it a second time, which the derivation does not call for.
For $x^2 + y^2 = 9$ (so $a = 3$), what is the total circle area by this chapter’s method?
$81 \pi$ — $81 = 3^4$, squaring the radius twice. The formula only squares $a$ once: $a^2 = 9$.
$3 \pi$ — The formula needs $a^2$, not $a$, $\pi \cdot 3^2 = 9 \pi$, not $\pi \cdot 3$.
$6 \pi$ — $6 \pi$ is the circumference, $2 \pi \cdot 3$. The area is $\pi a^2 = 9 \pi$.
The ellipse x²/9 + y²/4 = 1 inside the circle of radius 3. Every height is 2/3 of the circle’s, so its area is 6π.
the circle’s quarter is 9π/4
One quarter: 2/3 of the circle’s quarter, 9π/4, which is 3π/2. Four quarters make 6π ≈ 18.85.
Worked example
Find the area of the ellipse $x^2/9 + y^2/4 = 1$
$x^2/9 + y^2/4 = 1$, so $a = 3$, $b = 2$ Read $a$ and $b$ off the form $x^2/a^2 + y^2/b^2 = 1$.
$A = 4 \cdot (2/3) \int_0^3 \sqrt{9 - x^2} \mathrm{d} x$ Use the ellipse’s area: 4 times $b/a$ times the circle’s quarter, with $a = 3$ and $b = 2$.
$\int_0^3 \sqrt{9 - x^2} \mathrm{d} x = (\pi \cdot 3^2)/4 = (9 \pi)/4$ This integral is one quarter of the circle of radius 3.
*$A = 4 \cdot (2/3) \cdot (9 \pi)/4 = 6 \pi$ square units* Multiply. The answer is $\pi a b$ for $a = 3$ and $b = 2$.
Check: $0 < 6 \pi \approx 18.85 < 6 \times 4 = 24$ The answer is positive. It fits inside the rectangle 6 wide and 4 tall round the ellipse.
Region cut off by x equals 1: 8 over 3. Its rectangle, 1 wide and 4 tall: 4. It is two thirds of the rectangle.
Worked example
Find the area bounded by $y^2 = 4x$ and its latus rectum
$y^2 = 4x$, so $a = 1$; latus rectum $x = 1$ Match $y^2 = 4 a x$. Here $4 a = 4$, so $a = 1$, and the latus rectum is $x = a = 1$.
Upper arc $y = 2 \sqrt{x}$ Solve for $y$, and keep the top arc.
$A = 2 \int_0^1 2 \sqrt{x} \mathrm{d} x$ The $x$-axis is a mirror line. So double the area under the top arc.
$A = 2 \cdot [(4/3) x^{3/2}]_0^1$ An integral of $2 \sqrt{x}$ is $(4/3) x^{3/2}$. At 0 it is 0.
*$A = 2 \cdot (4/3) = 8/3$ square units* That is $(8/3) a^2$ for $a = 1$.
Check: $0 < 8/3 < 1 \times 4 = 4$ The answer is positive. The rectangle round the region is 1 wide and 4 tall, and the region is two thirds of it.
Check yourself
For $y^2 = 4x$ (so $a = 1$), the computed area bounded by the parabola and its latus rectum is \_\_\_\_\_\_.
$8/3$ square units
$128/3$ square units
$4/3$ square units
$16/3$ square units
Check your answer
✓ $8/3$ square units — (A) $(8/3) a^2 = (8/3) \cdot 1 = 8/3$ square units, matching the worked example’s evaluated integral.
$128/3$ square units — $4 a = 4$ gives $a = 1$, not $a = 4$. Using $a = 4$ by mistake gives $(8/3) \cdot 16 = 128/3$, the wrong quantity entirely.
$4/3$ square units — $4/3$ is missing the top-bottom symmetry factor of 2. The full formula is $(8/3) a^2 = 8/3$ when $a = 1$.
$16/3$ square units — The top-bottom symmetry factor of 2 is already fully accounted for inside $(8/3) a^2$. Doubling the result applies it a second time, which the derivation does not call for.
For $y^2 = 12 x$ (so $4 a = 12$, $a = 3$), what is the area bounded by the parabola and its latus rectum?
$8$ square units
$24$ square units
$72$ square units
$12$ square units
Check your answer
$8$ square units — The formula needs $a^2 = 9$, not $a = 3$, $(8/3) \cdot 3 = 8$ uses the wrong power of $a$.
$72$ square units — $72 = 8 \cdot 9$ skips dividing by 3. The formula is $(8/3) a^2 = 24$, not $8 a^2$.
$12$ square units — $12 = (4/3) \cdot 9$ is missing the top-bottom symmetry factor of 2. The full formula is $(8/3) a^2 = 24$.
One period of sin x: the arch above is 2, the arch below −2. Straight across gives 0; the area is 4.
Worked example
Find the area between $y = \sin(x)$ and the x-axis over one full period
$y = \sin(x)$, x-axis, $x = 0$ to $x = 2 \pi$ One full period of sine runs from 0 to $2 \pi$.
$\int_0^{2 \pi} \sin(x) \mathrm{d} x = [-\cos(x)]_0^{2 \pi} = 0$ This is one integral straight across. It gives 0, so it cannot be the area.
$\sin(x) > 0$ on $(0, \pi)$ and $\sin(x) < 0$ on $(\pi, 2 \pi)$ Sine crosses the axis at $x = \pi$. The arch above and the arch below cancel.
$\int_0^\pi \sin(x) \mathrm{d} x = [-\cos(x)]_0^\pi = 2$ Integrate the arch above the axis, from 0 to the crossing at $\pi$.
$\left|\int_\pi^{2 \pi} \sin(x) \mathrm{d} x\right| = \left|-2\right| = 2$ Integrate the arch below the axis on its own. Then take its size.
*$A = 2 + 2 = 4$ square units* Add the two sizes.
Check: $0 < 4 < 2 \pi \times 2 \approx 12.57$ The answer is positive. It fits inside the rectangle $2 \pi$ wide and 2 tall, from $y = -1$ to $y = 1$.
Check yourself
For $y = \sin x$ from $x = 0$ to $x = 2 \pi$, the true area, split at $x = \pi$, is \_\_\_\_\_\_.
$4$ square units
$-4$ square units
$2$ square units
$0$ square units
Check your answer
✓ $4$ square units — (A) Splitting at $x = \pi$: $\int_0^\pi \sin x d x = 2$ and $|\int_\pi^{2 \pi} \sin x d x| = 2$, so the true area is $2 + 2 = 4$.
$-4$ square units — Area is never negative. The second arch’s raw integral is $-2$, but its absolute value, $2$, is what gets added, giving a final, positive $4$.
$2$ square units — $2$ is the area of just ONE arch. Both the positive arch on $(0, \pi)$ and the negative arch on $(\pi, 2.\pi)$ contribute, giving $2 + 2 = 4$.
$0$ square units — $0$ is the raw, uncorrected integral, where the positive and negative arches cancel. Splitting and adding their absolute values gives $4$, not $0$.
Why does naively integrating $\sin x$ from $0$ to $2 \pi$ give $0$, while the true area is $4$?
the bounds $0$ and $2 \pi$ were chosen incorrectly
$\sin x$ is not actually integrable over a full period
the positive and negative arches exactly cancel each other
the two arches have different sizes, so one outweighs the other
Check your answer
the bounds $0$ and $2 \pi$ were chosen incorrectly — The bounds $0$ to $2 \pi$ are exactly as given. The fix is splitting the integral at the sign change within them, not choosing different bounds.
$\sin x$ is not actually integrable over a full period — $\sin x$ is continuous and integrable everywhere. The cancellation is a SIGN effect, not a failure of integrability.
✓ the positive and negative arches exactly cancel each other — (C) The raw integral lets the positive arch and the negative arch cancel exactly. Splitting at $x = \pi$ and taking the absolute value of each arch gives $2 + 2 = 4$ instead.
the two arches have different sizes, so one outweighs the other — The two arches are equal in size, that is exactly why they cancel EXACTLY, to $0$, rather than leaving some partial value behind.
For $y = \cos x$ between $x = 0$ and $x = \pi$, splitting at the sign change and taking absolute values, what is the TRUE area?
$1$ square unit
$2$ square units
$0$ square units
$4$ square units
Check your answer
$1$ square unit — $1$ is the area of just ONE half. Both the positive half on $(0, \pi/2)$ and the negative half on $(\pi/2, \pi)$ contribute, giving $1 + 1 = 2$.
✓ $2$ square units — (B) $\cos x$ changes sign at $x = \pi/2$. $\int_0^{\pi/2} \cos x d x = 1$ and $|\int_{\pi/2}^\pi \cos x d x| = 1$, so the true area is $1 + 1 = 2$.
$0$ square units — $0$ is the raw, uncorrected integral, where the positive and negative halves cancel. Splitting at $x = \pi/2$ and adding the absolute values gives $2$, not $0$.
$4$ square units — $4$ is the answer for the $\sin x$ example over the longer interval $[0, 2.\pi]$. Over the shorter interval $[0, \pi]$, $\cos x$’s two halves give $1 + 1 = 2$.
Six idea cards. One thin strip leads to the area under one curve, then to two curves, crossings, splitting, and the circle, ellipse and parabola.
the plot
line over parabola
split at 2
Three answers: the plot, 8.25; the line over the parabola between their meeting points, 9/8; the split curve, 8/3 + 16/3 = 8.
The first page asked a question. Can one subtraction, $F(b) - F(a)$, find the area of any region with a curved edge?
The three shapes: a circle, πa²; an ellipse, πab; a parabola to its latus rectum, (8/3)a². Here 4π, 2π and 8/3.
ONE CURVE: for $f(x) \geq 0$ on $[a, b]$, $A = \int_a^b f(x) \mathrm{d} x$. The plot under $y = x^2/4 + 1$ from 1 to 4 is 8.25.
FLAT STRIPS: for a curve $x = g(y)$, use horizontal strips: $A = \int_c^d g(y) \mathrm{d} y$.
TWO CURVES: if $f(x) \geq g(x)$ on $[a, b]$, $A = \int_a^b (f(x) - g(x)) \mathrm{d} x$.
NO LIMITS GIVEN: solve $f(x) = g(x)$. The line $y = 5x/4$ meets $y = x^2/4 + 1$ at 1 and 4.
A CROSSING: split wherever the integrand changes sign, at a crossing of the axis or of two curves. Then add the sizes. For $y = x^2/2 - 2$ on $[0, 4]$, that gives $8/3 + 16/3 = 8$.
THREE SHAPES: a circle $\pi a^2$, an ellipse $\pi a b$, and a parabola to its latus rectum $(8/3) a^2$.
the region
→
draw it
→
solve where
edges meet or cross
→
limits, crossings
limits, crossings
→
split
at every crossing
→
integrate
each piece, top minus bottom
→
add the sizes
no minus signs
→
the area
The four steps as a flow: draw, find the limits, split, integrate and add.
Giving the chapter its answer
Here is the answer. Yes, one method finds them all. First draw the region. Then find where its edges meet, since those points are the limits. Take one thin strip, and write its height as top minus bottom. Split wherever the integrand changes sign: where the curve crosses the axis, or where two curves cross. Add the strips in each piece with an integral, and add the sizes.
Three more regions on one axis: under one curve, 8/3; between two curves, 10/3; a quarter disc, π. The same method finds each.
The plot by the canal, the circle from Class 10, the ellipse and the parabola all go the same way.
Check yourself
Put this chapter’s area method in the order it is carried out.
add the sizes, integrate each piece, split at the crossings, find the bounds, sketch the region
sketch, find the bounds and every crossing, split there, integrate each piece top minus bottom, add the sizes
sketch, find the bounds, integrate top minus bottom straight across, and only afterwards split and take absolute values
integrate top minus bottom, sketch the region, find the bounds, split and add the sizes
Check your answer
add the sizes, integrate each piece, split at the crossings, find the bounds, sketch the region — This reverses the whole method. Adding the sizes is the last step, once each piece has been integrated.
✓ sketch, find the bounds and every crossing, split there, integrate each piece top minus bottom, add the sizes — (B) Sketch the region, find the bounds and every crossing, and split there. Then integrate each piece top minus bottom and add the sizes. Splitting first stops the pieces cancelling.
sketch, find the bounds, integrate top minus bottom straight across, and only afterwards split and take absolute values — Integrated straight across a crossing, the pieces have already cancelled. For $\sin x$ on $[0, 2 \pi]$ that gives $0$, not $4$. Split first, then integrate each piece.
integrate top minus bottom, sketch the region, find the bounds, split and add the sizes — Integrating needs the bounds and the top-minus-bottom setup already in place , sketching the region and finding the bounds must come first.
In the area method’s second step, where do the bounds come from when a problem does not give $x = a$, $x = b$ directly?
from the axis of symmetry of whichever curve is a parabola
from the derivative of the region’s area function
solving $f(x) = g(x)$: the curves’ own intersection
from the y-intercepts of the two curves
Check your answer
from the axis of symmetry of whichever curve is a parabola — The axis of symmetry belongs to one curve alone. The bounds instead come from where BOTH curves agree, via $f(x) = g(x)$.
from the derivative of the region’s area function — Differentiating an area function has no role here, the bounds come from where the two given curves physically meet, solved by $f(x) = g(x)$.
✓ solving $f(x) = g(x)$: the curves’ own intersection — (C) The area method’s bound-finding step is exactly this rule: when bounds are not given, solve $f(x) = g(x)$ and use the solutions as the bounds.
from the y-intercepts of the two curves — Y-intercepts only show each curve’s value at $x = 0$, they do not locate where the two curves cross each other elsewhere.
When can the split step be skipped?
when the bounds are given, so there is no crossing to find
when the region is not bounded by a circle or a parabola
when the integrand never changes sign on the interval
never, even where the integrand keeps its sign
Check your answer
when the bounds are given, so there is no crossing to find — Given bounds fix where the region starts and ends, not whether the integrand changes sign inside. $2 + x$ and $2 + x^3$ on $[-1, 1]$ cross at $0$, so that integral still needs the split.
when the region is not bounded by a circle or a parabola — The splitting step is about SIGN, not shape, it applies (or does not apply) based on whether the integrand changes sign, regardless of whether a circle or parabola is involved.
✓ when the integrand never changes sign on the interval — (C) The split fixes a sign change. If one curve stays on one side of the axis, or two curves do not cross inside the interval, the integrand keeps its sign and the plain integral is the area.
never, even where the integrand keeps its sign — The split step is a fix for a specific problem, a sign change. Where there is no sign change, the plain integral is already the true area, and no split is needed.
Which step of this chapter’s area method is the one borrowed whole from Chapter 7, with nothing new added?
integrating the top-minus-bottom difference
sketching or reasoning about the region
splitting the integral where the picture crosses
finding the bounds
Check your answer
✓ integrating the top-minus-bottom difference — (A) The integration itself, an antiderivative and $F(b) - F(a)$, comes whole from Chapter 7. Top minus bottom is new here, and splitting at a point $c$ is a Chapter 7 property put to a new use, area.
sketching or reasoning about the region — Sketching or reasoning about a region is new here, Chapter 7 dealt with integrals given directly, not with reading a picture to set one up.
splitting the integral where the picture crosses — Splitting an integral at a point $c$ is a Chapter 7 property. Using it for area, split where the sign changes and add the sizes, is new in this chapter.
finding the bounds — Finding the bounds, whether given, or solved from $f(x) = g(x)$, is new to this chapter. Chapter 7 always started with the bounds already fixed.
A student sketches the region, integrates bottom-minus-top by mistake, and reports a negative area with no further steps. Which step of the area method did the student get wrong?
the sketching step; the region was never drawn correctly
the bound-finding step; the wrong bounds were used
a different formula should have been used, since the region is bounded by a circle
reversed the top-minus-bottom order and left the crossings unchecked
Check your answer
the sketching step; the region was never drawn correctly — The stem says the student sketched the region, there is no indication the sketch itself was wrong. The error is in the subtraction order, and the negative answer went unchecked.
the bound-finding step; the wrong bounds were used — Nothing in the stem points to the bounds being wrong. A negative reported area, with no further correction, is the signature of a reversed subtraction order left unchecked.
a different formula should have been used, since the region is bounded by a circle — The stem gives no indication the region is bounded by a circle, or that a different formula is needed. The actual error is the subtraction order, and the negative answer went unchecked.
✓ reversed the top-minus-bottom order and left the crossings unchecked — (D) The student subtracted bottom minus top. A negative answer points back to the order, or to a crossing left unsplit, and the student checked neither.
Check yourself: the whole chapter
A classmate says: “$\int_a^b f(x) d x$ is defined to be the area under the curve $y = f(x)$.” Is that accurate?
Yes: area is exactly what the definite integral computes, in every case.
Not quite: a number, equal to the area only when $f(x) \geq 0$ on $[a, b]$.
No: the integral and the area are two separate ideas that share a symbol.
Yes, but only when the bounds $a$ and $b$ are both positive numbers.
Check your answer
Yes: area is exactly what the definite integral computes, in every case. — The integral is a number that can be negative or zero. It equals the area only when $f(x) \geq 0$ on $[a, b]$.
✓ Not quite: a number, equal to the area only when $f(x) \geq 0$ on $[a, b]$. — (B) A definite integral always evaluates to a number. That number equals the area under $y = f(x)$ only when the curve stays at or above the axis on $[a, b]$.
No: the integral and the area are two separate ideas that share a symbol. — They are closely linked: this chapter builds every area formula by computing a definite integral.
Yes, but only when the bounds $a$ and $b$ are both positive numbers. — The sign of $a$ and $b$ does not decide this. What matters is whether $f(x)$ stays non-negative on $[a, b]$.
The formula $A = \int_a^b y d x$ comes from adding up thin vertical strips. What is the area of ONE such strip, at position $x$?
Just $d x$: the strip’s width, on its own.
Just $y$: the strip’s height, on its own.
$y d x$: the height times the width.
$y^2 d x$: the disc-method volume element.
Check your answer
Just $d x$: the strip’s width, on its own. — $d x$ alone is a width, a bare length. Multiply in the height to get an area.
Just $y$: the strip’s height, on its own. — $y$ alone is a height. Multiply by the width $d x$ to get an area.
✓ $y d x$: the height times the width. — (C) The strip is a thin rectangle: height $y = f(x)$, width $d x$. Its area is the product, $y d x$.
$y^2 d x$: the disc-method volume element. — $y^2 d x$ is the disc-method volume element, from a different topic. The area strip is $y d x$.
A region’s boundary is naturally written as $x = h(y)$, not $y = f(x)$. Which strip keeps the integral straightforward?
A horizontal strip: $\int h(y) d y$, over the given $y$-bounds.
A vertical strip: $\int h(y) d x$, over the given $y$-bounds.
A vertical strip, after first solving $x = h(y)$ for $y$ in terms of $x$.
Either strip works the same, since area never depends on the choice.
Check your answer
✓ A horizontal strip: $\int h(y) d y$, over the given $y$-bounds. — (A) A horizontal strip fits $x = h(y)$ directly: width $h(y)$, thickness $d y$, giving $\int h(y) d y$ over the $y$-bounds.
A vertical strip: $\int h(y) d x$, over the given $y$-bounds. — The differential must match the strip. A horizontal strip is measured in $d y$, not $d x$.
A vertical strip, after first solving $x = h(y)$ for $y$ in terms of $x$. — Solving for $y$ is extra work here. The curve is already given the way the horizontal strip needs it.
Either strip works the same, since area never depends on the choice. — The choice matters. Picking against the grain of the equation forces solving it the harder way first.
On $[a, b]$, curve $f$ sits below curve $g$, so $g(x) \geq f(x)$ throughout. Which integral gives the area between them?
$\int_a^b (f(x) - g(x)) d x$
$\int_a^b (f(x) + g(x)) d x$
$\int_a^b f(x) g(x) d x$
$\int_a^b (g(x) - f(x)) d x$
Check your answer
$\int_a^b (f(x) - g(x)) d x$ — Order matters: $g$ is on top here, so it must come first. $f - g$ gives a negative number, not an area.
$\int_a^b (f(x) + g(x)) d x$ — Adding gives the combined size of both functions, not the gap between them. Area between curves is a subtraction.
$\int_a^b f(x) g(x) d x$ — Multiplying two functions belongs to a different problem. The gap between curves is top minus bottom.
✓ $\int_a^b (g(x) - f(x)) d x$ — (D) $g$ sits above $f$ on $[a, b]$, so the area is top minus bottom: $\int_a^b (g(x) - f(x)) d x$.
A curve $y = f(x)$ dips below the x-axis somewhere inside $[a, b]$. To find the actual area, what must you do?
Nothing extra: integrate straight from $a$ to $b$; the sign takes care of itself.
Split the interval at the crossing, and add the absolute value of each piece.
Multiply the whole integral by $-1$, flipping its sign once.
Ignore the part below the axis; only the part above it counts as area.
Check your answer
Nothing extra: integrate straight from $a$ to $b$; the sign takes care of itself. — The raw integral can cancel itself where the curve turns negative. It stops equalling the true area right at the crossing.
✓ Split the interval at the crossing, and add the absolute value of each piece. — (B) Split the interval at the crossing, take the absolute value of each piece, then add them. Every piece then contributes its true size.
Multiply the whole integral by $-1$, flipping its sign once. — One sign flip only fixes a piece that is entirely negative. Splitting at the crossing is what makes each piece positive on its own.
Ignore the part below the axis; only the part above it counts as area. — The region below the axis has area too, and it is not zero. It must be added in as a positive amount, not dropped.
Each one below is the same idea you already use, just named differently.
A sale’s consumer saving. A shopper tracks how much is saved at every price point during a clearance sale. The maths: The region between the price a buyer would pay and the lower sale price is top curve minus bottom curve. Add that over every unit sold. A sale drops the price by ₹$20$ on each of $15$ units sold, a total saving of $20 \times 15 = 300$ rupees.
A tailor’s curved hemline. A tailor checks how much extra cloth a curved hem needs. The maths: The extra cloth is the area between the curved hem and the straight one. It is bounded between two ordinates. A straight hem needs $40$ square centimetres and the curved hem needs $58$, an extra $58-40=18$.
A streaming app’s hourly usage. An app totals how many hours of video stream each evening. The maths: Adding thin strips under that usage curve gives the evening’s total hours streamed. Each strip is a height times a width. Six hourly strips, each at a height of $0.5$, add to $6 \times 0.5 = 3$ total hours’ worth of streaming.
A badge’s gold face. A badge maker prices the gold leaf covering an oval badge’s whole face. The maths: A badge’s face is an ellipse. Its area is already pi times its two half-axes multiplied together. An oval badge with half-axes $4$ cm and $2$ cm has a gilded area of $\pi \times 4 \times 2 = 8 \pi$ square centimetres.
An airline’s fuel-burn record. An airline totals how much fuel one flight burns each minute. The maths: A burn rate added up minute by minute is the area under one curve. That curve is bounded between take-off and landing. A flight burns fuel at a steady $2000$ kg per hour for $3$ hours, a total of $2000 \times 3 = 6000$ kg.
A baker’s round cake tier. A baker works out how much icing covers a round cake tier. The maths: That tier’s top is a circle. One integrated quarter of it, multiplied by four, gives pi times its radius squared. A cake tier of radius $10$ cm has a top area of $\pi \times 10^2 = 100 \pi$ square centimetres.
Your turn: A tap fills a tank at a steady $5$ litres per minute for $12$ minutes. Using the area-under-a-rate idea, how many litres does it add?
LATER
Where this area idea goes next
First-year engineering mechanics reuses this strip-by-strip idea for a centroid, a flat shape’s average position, and for a moment of inertia, how hard that shape is to turn about an axis. Both replace a plain strip with one weighted by its own distance from an axis, over a region this chapter already bounds and splits.
Exercise 8.1 practises the area under one curve. The curves are $y = x^2$, $y = x^3$, a circle, a parabola and an ellipse. Draw each region and one strip before you integrate.
Exercise 8.1: Area under a curve
practice Find the area bounded by $y = x^2$, the x-axis, and the lines $x = 1$ and $x = 3$.
practice Find the area bounded by $y = x^3$, the x-axis, and the lines $x = 0$ and $x = 2$.
practice Find the area enclosed by the circle $x^2 + y^2 = 25$.
practice Find the area bounded by the parabola $y^2 = 16 x$ and its latus rectum.
practice Find the area enclosed by the ellipse $x^2/16 + y^2/4 = 1$.
practice Find the area of the region bounded by $x^2 = 4 y$, the lines $y = 2$ and $y = 4$, and the y-axis, in the first quadrant.
practice Find the area of the region in the first quadrant under the ellipse $x^2/16 + y^2/9 = 1$, between $x = 0$ and $x = 2$.
practice Find the area of the region bounded by $y^2 = 9 x$, the lines $x = 2$ and $x = 4$, and the x-axis, in the first quadrant.
These questions practise the area between two curves. Each pair is a line and a curve, or two parabolas. Find where the curves meet, then integrate top minus bottom between those points.
Area between two curves
practice Find the area between $y = x$ and $y = x^3$, for $x \geq 0$.
practice Find the area between $y = 4$ and $y = x^2$.
practice Find the area between $y = x^2 + 1$ and $y = 2 x + 1$.
practice Find the area enclosed between the parabolas $y^2 = 4 x$ and $x^2 = 4 y$.
practice Find the area of the region bounded by the parabola $x^2 = 4 y$ and the line $y = 4$.
The Miscellaneous Exercise mixes everything you have met. Some curves dip below the axis, so check where each one crosses before you integrate. Then decide whether the area needs a split.
Miscellaneous: the whole chapter
practice Find the area of the part of the circle $x^2 + y^2 = 36$ that lies in the first quadrant.
practice Find the area bounded by $y = x^3$, the x-axis, and the lines $x = -1$ and $x = 1$.
practice Find the area bounded by $y = x^2 - 4$ and the x-axis, between $x = -2$ and $x = 2$.
practice Find the area of the region enclosed between the two arcs of $y^2 = x$ and the lines $x = 1$ and $x = 4$.
practice Find the area of the region bounded by $y = x^2$ and $y = \left|x\right|$. (JEE tier)
Answers
$9 \pi$ square units
$1/2$ square units
$32/3$ square units
$28/3$ square units
$1/3$ square unit
Board paper practice
practice The area bounded by $y = \sin x$, the x-axis, $x = 0$ and $x = \pi$ is ______. (1 mark)
$1$ square unit
$2$ square units
$\pi/2$ square units
$\pi$ square units
practice Assertion (A). The area enclosed by the ellipse $x^2/25 + y^2/4 = 1$ is $10 \pi$ square units. Reason (R). For an ellipse $x^2/a^2 + y^2/b^2 = 1$, the area is $\pi a b$. (1 mark)
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.
practice Find the area bounded by $y = 9 - x^2$ and the x-axis. (2 marks)
practice Find $k > 0$ such that the area bounded by $y = k x^2$, the x-axis, $x = 0$ and $x = 2$ is $40/3$ square units. (3 marks)
practice Find the area of the smaller region cut off from the circle $x^2 + y^2 = 16$ by the line $x = 2 \sqrt{2}$. (5 marks)
practice A car’s speed is $v(t) = 60 - 10 t$ km/h at time $t$ hours, for $0 \leq t \leq 6$. (i) Find the distance covered in the first $2$ hours. (1 mark) (ii) Find the total distance covered in $6$ hours. (1 mark) (iii) Find the car’s average speed over those $6$ hours. (2 marks)
practice Find the area bounded by $y = x^3 - 4 x$ and the x-axis, between $x = -2$ and $x = 2$. (3 marks)
Answers
(B) $2$ square units.
(A) Both A and R are true and R is the correct explanation of A ($10 \pi$ square units).
