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Undoing the derivative
Many homes have a water tank on the roof. A pipe fills it from below. Say you know how fast the water flows in, in litres a minute. But you cannot see inside the tank.
From the flow alone, can you tell how much water came in during ten minutes? Can you tell how much is in the tank now?
Chapters 5 and 6 went one way. You started from a quantity and found its rate of change. Here is the question to carry to the last page. Can you go back, from the rate to the quantity?
Going back from a rate is called INTEGRATION. It is the opposite of a derivative: an ANTIDERIVATIVE. In this chapter you first go back from a rate to a quantity. Then you turn a rate into one number. That number is the total change between two moments. After that come the methods. Each one is a rule from Chapter 5, read backwards.
The tank comes back several times. Keep it in mind.
Kanada lived in north India between the 6th and the 2nd century BCE. His school, the Vaisheshika, sorted all that exists into kinds. He taught that a large thing is built from very small parts, joined together. Ask the tank Kanada’s question: what is its water made of? Each minute of flow adds one small part. Going back from the rate to the amount means adding those parts up.
- What does integrating a function mean, most directly?
Check your answer
- $x^2$ and $x^2+5$ share the same derivative, $2x$. What does this illustrate about integration?
Check your answer
- How does a definite integral differ from an indefinite integral, as this chapter frames it?
Check your answer
Before you start
You have already used a tool like each one below, long before now. Try each one below.
- Derivative of a polynomial (Class 11, Mathematics (NCERT Class 11), page 246).
Here: Running that rule backwards builds the standard integrals table. One derivative, read in reverse, becomes one table entry.
Check: the term $x^3$ differentiates to $3x^2$, and at $x=2$ this gives $3(2)^2=12$.
- Division algorithm for polynomials (Class 9, Mathematics, Polynomials).
Here: An improper fraction here needs exactly that division done first; the worked example Dividing first in Partial fractions before integrating, later in this chapter, shows one.
Check: dividing $x^2+1$ by $x-1$ gives $x^2+1 = (x-1)(x+1) + 2$, a quotient $x+1$ and a remainder $2$.
- Solving a pair of linear equations (Class 10, Mathematics (NCERT Class 10), page 30).
Here: Matching coefficients for A and B here solves two linear equations in two unknowns, exactly this method again.
Check: solving $A(x+2)+B(x-1)=3x+3$ at $x=1$ gives $3A=6$, so $A=2$. At $x=-2$ it gives $-3B=-3$, so $B=1$.
- Completing the square in a quadratic (Class 11, Mathematics, Conic Sections, page 180).
Here: Turning $x^2-6x+5$ into $(x-3)^2-4$ here produces the exact shape this toolkit matches against, aimed at a denominator instead of a quadratic equation.
Check: completing the square on $x^2-6x+5$ gives $(x-3)^2-4$, matching $3^2-4=5$ and $-2(3)=-6$.
- Closed interval (Class 11, Mathematics (NCERT Class 11), page 11).
Here: Splitting $[a,b]$ into strips here uses exactly this same closed interval, both endpoints included throughout.
Check: the closed interval $[2,6]$ keeps both endpoints, and splits into $n=4$ equal strips of width $h=(6-2)/4=1$.
If any of these felt new, read the page named before going on.
Integrals come in families
Go back to the tank. Say the water flows in at a steady 20 litres a minute. After 5 minutes, 100 litres have come in. If the tank held 100 litres at the start, it now holds 200. If it held 300, it now holds 400. In both tanks the water rises at the same rate.
So the flow tells you how the amount changes. It does not tell you where it started. Every starting amount fits the same flow.
A function $F$ with $F'(x) = f(x)$ is an INTEGRAL of $f$. You write the whole family of integrals as
$\int f(x) \mathrm{d} x = F(x) + C$.
This is the INDEFINITE INTEGRAL of $f$. The function $f$ is the INTEGRAND. The number $C$ is the CONSTANT OF INTEGRATION. Each value of $C$ is one starting amount.
For the tank, the flow is $f(t) = 20$, so $\int 20 \mathrm{d} t = 20 t + C$. With $C = 100$ you get the first tank. With $C = 300$ you get the second.
Why can two integrals of the same function differ only by a constant? Say $F$ and $G$ both have slope $f$ on an interval. Then $F - G$ has slope 0 at every point of it. A function with slope 0 all along an interval is constant there. Chapter 5’s mean value theorem is the reason.
Let us check it on $x^2$ and $x^2 + 5$. Differentiate each one:
$d/d x (x^2) = 2 x$,
$d/d x (x^2 + 5) = 2 x$.
The two differ by 5 at every $x$, yet both have slope $2 x$. So each is an integral of $2 x$, and the $+ C$ covers the 5:
$\int 2 x \mathrm{d} x = x^2 + C$.
So never drop the $C$. Without it you have one tank, not all of them.
Your turn: differentiate $(x + 1)^2$ and $x^2 + 2 x$. Both are integrals of the same function. How far apart are they?
Two tanks on one flow. What makes this tank this tank, and not the other? Both take in $20$ litres a minute, and both rise along $20 t + C$. The one thing that tells them apart is where each one started, $C = 100$ or $C = 300$. It is the one fact the flow cannot give you.
- The indefinite integral $\int f(x) d x$ denotes \_\_\_\_\_\_.
Check your answer
- Why do two integrals of the same $f$ differ only by a constant?
Check your answer
- $\int \cos x d x = \sin x + C$. Which of these IS a member of this integral family?
Check your answer
- Given $f'(x) = \cos x$ and $f(0) = 3$, what determines the value of $C$ in $f(x) = \sin x + C$?
Check your answer
The definite integral as a limit of totals
Counting squares for area
In Class 6 or 7 you may have traced a leaf on squared paper. To find its area, you counted the squares inside the outline. A full square counted as one. A square more than half inside counted as one too, and the rest were left out. With smaller squares, the count came closer to the true area.
So you can find an area by adding many thin pieces. The thinner the pieces, the closer the total.
The same idea gives the area under a graph. Let $f$ be continuous on $[a, b]$, with its graph above the axis. Split $[a, b]$ into $n$ equal strips, each of width $h = (b - a)/n$. On each strip, stand a rectangle as tall as $f$ at its left end. Add up the rectangles. As $n$ grows, the strips get thinner and the total settles on one number. That number is the DEFINITE INTEGRAL of $f$ from $a$ to $b$.
$\int_a^b f(x) \mathrm{d} x = \lim_{n \to \infty} h [f(a) + f(a + h) + \cdots + f(a + (n - 1) h)]$.
The figure does this for $x^2/4 + 1$ on $[0, 4]$. With 4 strips the total is 7.5. With 8 strips it is 8.375, and with 16 it is 8.84375. The totals climb towards $28/3$, about 9.33.
The board exam does not set this section. It shows what a definite integral is. NCERT’s name for it is the limit of a sum. The fundamental theorem, in the next section, gives a much quicker way to find one.
What are the smallest parts of an area? Ask it of the strips. Is there a thinnest strip, where you can stop? Each finer split gives a new total, and none of them is the last. So the area is not any one of these totals. It is the number they settle on, $28/3$.
- Before the Fundamental Theorem shortcut, $\int_a^b f(x) d x$ is defined as \_\_\_\_\_\_.
Check your answer
- In the total $h[f(a)+f(a+h)+\cdots+f(a+(n-1)h)]$, what is $h$?
Check your answer
- As $n \to \infty$ in the Riemann sum, what happens, and why does it matter?
Check your answer
The fundamental theorem of calculus
Go back to the tank, and draw its flow against time. Say the flow is a steady 20 litres a minute. The graph is a flat line at height 20. In 5 minutes, 100 litres come in. That is also the area under the line, a rectangle 20 tall and 5 wide. So the water so far is the area under the flow graph so far.
Now look at the next minute. The tank gains about one minute’s flow. So the water so far grows at the rate of the flow.
The same holds for any graph. Let $f$ be continuous on $[a, b]$. The AREA FUNCTION
$A(x) = \int_a^x f(t) \mathrm{d} t$
is the area under $f$ from $a$ up to $x$. The FIRST FUNDAMENTAL THEOREM says that $A'(x) = f(x)$ for every $x$ in $[a, b]$.
Here is why. Move $x$ on by a small step $h$. The area gains a thin strip, about $f(x)$ tall and $h$ wide. So
$A(x + h) - A(x) \approx f(x) \cdot h$.
Divide both sides by $h$:
$(A(x + h) - A(x))/h \approx f(x)$.
*Now let $h$* shrink to 0. The left side becomes the derivative of $A$, as in Chapter 5, and the strip fits better and better:
$A'(x) = f(x)$.
In the figure, the curve is 2.125 tall at $x = 3$. A strip 0.5 wide adds about $2.125 \cdot 0.5 = 1.0625$.
Check it with $f(t) = 2 t$ from 0. The area up to $x$ is a triangle, $x$ wide and $2 x$ tall. Half of $x$ times $2 x$ gives
$A(x) = x^2$.
Differentiate it, as in Chapter 5:
$A'(x) = 2 x$, which is $f(x)$.
So the area function is an integral of $f$. Finding areas and finding integrals are one job.
- For $A(x) = \int_a^x f(t) d t$, what does the first fundamental theorem state?
Check your answer
- What is $A(x) = \int_a^x f(t) d t$, conceptually?
Check your answer
- Why is the first fundamental theorem the deep reason integration and differentiation are inverse operations?
Check your answer
Many homes also have a water meter on the pipe. It keeps a running total of the litres that have passed. Say it reads 1150 litres at 6:00 and 1350 litres at 6:10. Then 200 litres came in between 6:00 and 6:10. You did not add up the flow minute by minute. Two readings were enough. And it does not matter how long ago the meter started counting.
So the total change over a stretch is the reading at the end minus the reading at the start.
The SECOND FUNDAMENTAL THEOREM says this in symbols. Let $f$ be continuous on $[a, b]$, and let $F$ be any integral of $f$. Then
$\int_a^b f(x) \mathrm{d} x = F(b) - F(a)$.
The short way to write $F(b) - F(a)$ is $[F(x)]_a^b$. Work out $F$ at the top limit, then take away its value at the bottom limit. So
$\int_a^b f(x) \mathrm{d} x = [F(x)]_a^b$.
The constant cancels. With $F + C$ in place of $F$, you get
$(F(b) + C) - (F(a) + C) = F(b) - F(a)$.
So a definite integral is one number, with no $+ C$.
For the tank, the flow is 20 litres a minute, and $F(t) = 20 t + C$. Put in the end time 10 and the start time 0, and subtract:
$F(10) - F(0) = (200 + C) - (0 + C)$.
*The $C$ cancels* again:
$F(10) - F(0) = 200$.
That is the 200 litres on the meter.
Let us find $\int_0^3 2 x \mathrm{d} x$ with two different integrals.
Find $\int_0^3 2 x \mathrm{d} x$ with two integrals
- $F(x) = x^2$
An integral of $2 x$, since the derivative of $x^2$ is $2 x$. - $\int_0^3 2 x \mathrm{d} x = F(3) - F(0)$
The second fundamental theorem: the value at the top limit minus the value at the bottom. - *$F(3) - F(0) = 9 - 0 = 9$*
Work out each value, then subtract. - $(9 + 5) - (0 + 5) = 14 - 5 = 9$
Check with $F(x) = x^2 + 5$: the 5 cancels, and the answer is 9 again.
So a definite integral needs one integral, $F$, and two values of it.
Your turn: find $\int_0^3 (2 x - 2) \mathrm{d} x$. Then sketch $y = 2 x - 2$. Where is the graph below the axis?
- The second fundamental theorem states that for any integral $F$ of $f$ on $[a,b]$, $\int_a^b f(x) d x$ equals \_\_\_\_\_\_.
Check your answer
- What role does the second fundamental theorem play in this chapter?
Check your answer
- Why does the constant of integration never appear in a definite-integral answer?
Check your answer
- To evaluate $\int_1^2 (3x^2-2x) d x$ using the second fundamental theorem, what is the first step?
Check your answer
A definite integral via the Fundamental Theorem: $\int_1^2 (3x^2-2x) \mathrm{d} x$
- Find $\int_1^2 (3x^2 - 2x) \mathrm{d} x$
The integrand is continuous on $[1, 2]$, so the second fundamental theorem applies. First find one integral. - $F(x) = x^3 - x^2$
Differentiate to test it. $F'(x) = 3 x^2 - 2 x$, the integrand. - $\int_1^2 (3x^2-2x) \mathrm{d} x = F(2) - F(1)$
Subtract the value at the lower limit from the value at the upper limit. - $F(2)-F(1) = (8-4)-(1-1) = 4-0 = 4$
Work out each value, then subtract. - $\int_1^2 (3x^2-2x) \mathrm{d} x = 4$
One number, with no $+ C$. - $F(x) = x^3 - x^2 + 7$ gives $(8 - 4 + 7) - (1 - 1 + 7) = 11 - 7 = 4$
Check with another integral. The 7 cancels, and the answer is 4 again.
Find the reversible isothermal work done by an ideal gas, using the Fundamental Theorem
- $w_\text{rev} = -\int_{V_i}^{V_f} p \mathrm{d} V$
Chemistry 11, Chapter 5, Thermodynamics, defines reversible work this way, the gas expanding or compressing against pressure $p$ from volume $V_i$ to volume $V_f$. - for $n$ mol of an ideal gas at constant temperature $T$, $p = (n R T)/V$
The ideal gas law, with temperature fixed throughout this isothermal process. - $w_\text{rev} = -n R T \int_{V_i}^{V_f} 1/V \mathrm{d} V$
Substitute $p$; $n$, $R$ and $T$ are all constants, so they come straight out of the integral. - $F(V) = \ln V$
The integral of $1/V$ is $\ln V$, as $1/x$ gives $\ln x$ in Chapter 5; the same pattern as the FTC example above; no substitution is needed here. - $\int_{V_i}^{V_f} 1/V \mathrm{d} V = F(V_f)-F(V_i) = \ln V_f - \ln V_i$
The second fundamental theorem. - *$w_\text{rev} = -n R T (\ln V_f - \ln V_i) = -n R T \ln(V_f/V_i)$*
Combine the two logs into one, matching the formula Chemistry 11 already states.
- For $\int_1^2 (3x^2-2x) d x$, an integral is $F(x)$ equal to \_\_\_\_\_\_.
Check your answer
- With $F(x)=x^3-x^2$, $F(2)-F(1)$ equals \_\_\_\_\_\_.
Check your answer
- Why does this worked example need no limiting Riemann sum?
Check your answer
Here is a working for the area between $y = x$ and the $x$-axis, from $x = -1$ to $x = 2$. Which step is wrong?
- $\int_{-1}^2 x \mathrm{d} x = [x^2/2]_{-1}^2 = 2 - 0.5 = 1.5$.
- So the area is 1.5.
Keep your answer, then look at the graph. On $[-1, 0]$ the line is below the axis. That piece has area 0.5, but the integral counts it as $-0.5$. On $[0, 2]$ the line is above the axis, with area 2. The integral adds the two pieces with their signs:
$-0.5 + 2 = 1.5$.
The area adds their sizes:
$0.5 + 2 = 2.5$.
The second step is wrong. A definite integral counts a piece below the axis as negative. In the tank, water flowing out counts against the water flowing in.
To find an area, split the interval at every point where the graph crosses the axis. Then add the sizes of the pieces.
You called the integral and the area, $1.5$ and $2.5$ from the last page, both “the area”. Are they the same kind of thing? The integral is a signed total. It counts the piece below the axis as $-0.5$, as the tank counts water going out. An area is a size, and a size is never below $0$. So on $[-1, 2]$ the integral is $1.5$, but the area is $2.5$. Before you answer, ask which kind of number the question wants.
- The claim ‘the definite integral $\int_a^b f(x) d x$ IS the area under the curve’ is \_\_\_\_\_\_.
Check your answer
- $\int_{-1}^2 x d x = 1.5$. What is the true AREA between $y=x$ and the $x$-axis on $[-1,2]$?
Check your answer
- Why does $f$ crossing zero inside $[a,b]$ break the ’integral equals area’ claim?
Check your answer
- To find the true area when $f$ crosses zero inside $[a,b]$, the fix is to \_\_\_\_\_\_.
Check your answer
The standard integrals table
The Fundamental Theorem turned an area into one job: find an integral. This table does that job for the functions you know from Chapter 5. Three methods follow to do the rest. Substitution is the chain rule read backwards. By parts is the product rule read backwards. Partial fractions is adding fractions, undone. Between them sits a table of nine formulas found by those methods, kept for lookup.
Each formula in this table is a Chapter 5 derivative, read from right to left. You know that
$d/d x (\sin x) = \cos x$.
*So $\sin x$* is an integral of $\cos x$:
$\int \cos x \mathrm{d} x = \sin x + C$.
The power rule needs one condition. For $n \neq -1$,
$\int x^n \mathrm{d} x = x^{n+1}/(n+1) + C$.
At $n = -1$ this would divide by 0. That case has its own formula. For $x \neq 0$, $\int 1/x \mathrm{d} x = \ln|x| + C$.
The other formulas come the same way. You do not learn this table now. Each line is one Chapter 5 derivative read backwards, and you look it up when you need it.
| You know, from Chapter 5 | So |
|---|---|
| $d/d x (e^x) = e^x$ | $\int e^x \mathrm{d} x = e^x + C$ |
| $d/d x (a^x) = a^x \ln a$ | $\int a^x \mathrm{d} x = a^x/(\ln a) + C$, for $a > 0$, $a \neq 1$ |
| $d/d x (-\cos x) = \sin x$ | $\int \sin x \mathrm{d} x = -\cos x + C$ |
| $d/d x (\sin x) = \cos x$ | $\int \cos x \mathrm{d} x = \sin x + C$ |
| $d/d x (\tan x) = \sec^2 x$ | $\int \sec^2 x \mathrm{d} x = \tan x + C$ |
| $d/d x (-\cot x) = \text{cosec}^2 x$ | $\int \text{cosec}^2 x \mathrm{d} x = -\cot x + C$ |
| $d/d x (\sec x) = \sec x \tan x$ | $\int \sec x \tan x \mathrm{d} x = \sec x + C$ |
| $d/d x (-\text{cosec} x) = \text{cosec} x \cot x$ | $\int \text{cosec} x \cot x \mathrm{d} x = -\text{cosec} x + C$ |
| $d/d x (\sec^{-1} x) = 1/(x \sqrt{x^2 - 1})$ | $\int 1/(x \sqrt{x^2 - 1}) \mathrm{d} x = \sec^{-1} x + C$, for $x > 1$ |
Two rules let you use the formulas together. The integral of a sum is the sum of the two integrals. A constant factor comes outside, so $\int k f(x) \mathrm{d} x = k \int f(x) \mathrm{d} x$.
Check one formula by differentiating. Take $\int \text{cosec}^2 x \mathrm{d} x = -\cot x + C$. The derivative of $\cot x$ is $-\text{cosec}^2 x$, so
$d/d x (-\cot x) = -(-\text{cosec}^2 x)$.
Two minus signs make a plus:
$d/d x (-\cot x) = \text{cosec}^2 x$.
That is the integrand, so the formula is right.
So you can check every answer the same way. Differentiate it, and you should get back the integrand.
- For $n \neq -1$, $\int x^n d x$ equals \_\_\_\_\_\_.
Check your answer
- Why does the power rule $\int x^n d x = x^{n+1}/(n+1)+C$ exclude $n=-1$?
Check your answer
- $\int e^x d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int \sec^2 x d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int (x^3+2x)/x d x$ needs which step before a table entry applies?
Check your answer
Integration by substitution
The chain rule from Chapter 5 is for a composite function, one function applied to the output of another. Their slopes multiply. So if $F' = f$, then
$d/d x (F(g(x))) = f(g(x)) \cdot g'(x)$.
Read it backwards: $F(g(x))$ is an integral of the right side. That is INTEGRATION BY SUBSTITUTION.
$\int f(g(x)) \cdot g'(x) \mathrm{d} x = F(g(x)) + C$.
To use it, put
$t = g(x)$.
Differentiate both sides with respect to $x$:
$d t/d x = g'(x)$.
Then write it with $\mathrm{d} t$ and $\mathrm{d} x$ apart. This is shorthand: wherever the integral has $g'(x) \mathrm{d} x$, write $\mathrm{d} t$. The chain rule above is what allows it.
$\mathrm{d} t = g'(x) \mathrm{d} x$.
So the integral becomes $\int f(t) \mathrm{d} t$. Integrate in $t$. Then put $g(x)$ back in place of $t$.
So look for a function whose derivative also appears in the integrand. It may appear there times a constant, and that is fine.
Let us try $\int 2 x \cos(x^2) \mathrm{d} x$, one line at a time.
Find $\int 2 x \cos(x^2) \mathrm{d} x$ by substitution
- $t = x^2$
The derivative of $x^2$ is $2 x$, and $2 x$ appears in the integrand. - $\mathrm{d} t = 2 x \mathrm{d} x$
Differentiate $t = x^2$, then write $\mathrm{d} t$ in place of $2 x \mathrm{d} x$. - $\int 2 x \cos(x^2) \mathrm{d} x = \int \cos t \mathrm{d} t$
Put in $t$ and $\mathrm{d} t$: now the integral is in $t$ alone. - $\int \cos t \mathrm{d} t = \sin t + C$
The table gives the integral of $\cos t$. - *$\int 2 x \cos(x^2) \mathrm{d} x = \sin(x^2) + C$*
Put $x^2$ back in place of $t$, so the answer is in $x$. - $d/d x (\sin(x^2)) = \cos(x^2) \cdot 2 x$
Check by differentiating with the chain rule: the integrand comes back.
The answer must be in $x$, the variable you started with. An answer left in $t$ is not finished.
Your turn: find $\int 2 x (x^2 + 3)^4 \mathrm{d} x$. Which $t$ turns it into a power of $t$?
- Substitution works by setting $t=g(x)$ to turn $\int f(g(x)) \cdot g'(x) d x$ into \_\_\_\_\_\_.
Check your answer
- What should you look for in the integrand to spot a good substitution $t=g(x)$?
Check your answer
- After integrating $\int \sqrt{t} d t$ during a substitution, what must happen before the answer is complete?
Check your answer
- $\int 2x \cos(x^2) d x$ needs substitution, but $\int \cos x d x$ does not. What is the difference?
Check your answer
Substitution, start to finish: $\int 2x \sqrt{x^2+1} \mathrm{d} x$
- Find $\int 2x \sqrt{x^2+1} \mathrm{d} x$
Look for a function whose derivative also appears in the integrand. - $t = x^2+1$, so $\mathrm{d} t = 2x \mathrm{d} x$
The derivative of $x^2 + 1$ is $2 x$, and $2 x$ is already there. - $\int 2x \sqrt{x^2+1} \mathrm{d} x = \int \sqrt{t} \mathrm{d} t$
Put in $t$ and $\mathrm{d} t$. Now the integral is in $t$ alone. - $\int \sqrt{t} \mathrm{d} t = 2/3 \cdot t^{3/2} + C$
This is the power rule, with $n = 1/2$. - $\int 2x \sqrt{x^2+1} \mathrm{d} x = 2/3 \cdot (x^2+1)^{3/2} + C$
Put $x^2 + 1$ back in place of $t$. The answer must be in $x$. - Check by differentiating: $d/d x (2/3 \cdot (x^2+1)^{3/2}) = 2/3 \cdot 3/2 \cdot (x^2+1)^{1/2} \cdot 2x = 2x \sqrt{x^2+1}$
Check by differentiating the answer with the chain rule. You get back the integrand.
Four more formulas come from substitution, for $\tan x$, $\cot x$, $\sec x$ and $\text{cosec} x$. Here is the one for $\tan x$, line by line.
Find $\int \tan x \mathrm{d} x$
- $\int \tan x \mathrm{d} x = \int (\sin x)/(\cos x) \mathrm{d} x$
Write $\tan x$ as $(\sin x)/(\cos x)$. - $t = \cos x$, so $\mathrm{d} t = -\sin x \mathrm{d} x$
The top is minus the derivative of the bottom, so put $t$ for the bottom. - $\int (\sin x)/(\cos x) \mathrm{d} x = -\int (\mathrm{d} t)/t$
Write $-\mathrm{d} t$ in place of $\sin x \mathrm{d} x$. - $-\int (\mathrm{d} t)/t = -\ln \left|\cos x\right| + C$
The integral of $1/t$ is $\ln \left|t\right|$, then put $\cos x$ back. - *$\int \tan x \mathrm{d} x = \ln \left|\sec x\right| + C$*
A minus sign before a log turns $\cos x$ into $1/(\cos x)$, which is $\sec x$.
The cot formula is the same, with $t = \sin x$: $\int \cot x \mathrm{d} x = \ln \left|\sin x\right| + C$.
The sec formula needs one more step. Multiply above and below by $\sec x + \tan x$.
Find $\int \sec x \mathrm{d} x$
- $\int \sec x \mathrm{d} x = \int (\sec x (\sec x + \tan x))/(\sec x + \tan x) \mathrm{d} x$
Multiply above and below by $\sec x + \tan x$, which changes nothing. - $\int \sec x \mathrm{d} x = \int (\sec^2 x + \sec x \tan x)/(\sec x + \tan x) \mathrm{d} x$
Open the bracket on top. - $t = \sec x + \tan x$, so $\mathrm{d} t = (\sec x \tan x + \sec^2 x) \mathrm{d} x$
The top is the derivative of the bottom, so put $t$ for the bottom. - $\int (\mathrm{d} t)/t = \ln \left|t\right| + C$
The integral of $1/t$ is $\ln \left|t\right|$. - *$\int \sec x \mathrm{d} x = \ln \left|\sec x + \tan x\right| + C$*
Put $\sec x + \tan x$ back in place of $t$.
The cosec formula works the same way, with $\text{cosec} x - \cot x$ above and below: its derivative is $\text{cosec} x (\text{cosec} x - \cot x)$, the new top. $\int \text{cosec} x \mathrm{d} x = \ln \left|\text{cosec} x - \cot x\right| + C$.
- For $\int 2x \sqrt{x^2+1} d x$, the natural substitution is \_\_\_\_\_\_.
Check your answer
- With $t=x^2+1$, $\int 2x \sqrt{x^2+1} d x$ becomes \_\_\_\_\_\_.
Check your answer
- $\int \sqrt{t} d t = 2/3 \cdot t^{3/2} + C$. Substituting back, $\int 2x \sqrt{x^2+1} d x$ equals \_\_\_\_\_\_.
Check your answer
A toolkit of special-form integrals
Some integrals come up again and again. For those you keep the formula and look it up. There are nine. You match the shape and read off the answer. In every formula $a > 0$. Each answer holds wherever its integrand is defined. Differentiate any answer and you get its integrand back. That is the check.
The nine are in one table, in three groups of three.
| The integral | The answer |
|---|---|
| No square root | |
| $\int 1/(x^2 + a^2) \mathrm{d} x$ | $(1/a) \tan^{-1}(x/a) + C$ |
| $\int 1/(x^2 - a^2) \mathrm{d} x$ | $(1/(2a)) \ln|(x - a)/(x + a)| + C$ |
| $\int 1/(a^2 - x^2) \mathrm{d} x$ | $(1/(2a)) \ln|(a + x)/(a - x)| + C$ |
| A square root below the line | |
| $\int 1/\sqrt{a^2 - x^2} \mathrm{d} x$ | $\sin^{-1}(x/a) + C$ |
| $\int 1/\sqrt{x^2 + a^2} \mathrm{d} x$ | $\ln|x + \sqrt{x^2 + a^2}| + C$ |
| $\int 1/\sqrt{x^2 - a^2} \mathrm{d} x$ | $\ln|x + \sqrt{x^2 - a^2}| + C$ |
| The square root on top | |
| $\int \sqrt{a^2 - x^2} \mathrm{d} x$ | $(x/2) \sqrt{a^2 - x^2} + (a^2/2) \sin^{-1}(x/a) + C$ |
| $\int \sqrt{x^2 + a^2} \mathrm{d} x$ | $(x/2) \sqrt{x^2 + a^2} + (a^2/2) \ln|x + \sqrt{x^2 + a^2}| + C$ |
| $\int \sqrt{x^2 - a^2} \mathrm{d} x$ | $(x/2) \sqrt{x^2 - a^2} - (a^2/2) \ln|x + \sqrt{x^2 - a^2}| + C$ |
To prove one formula, take $\int 1/\sqrt{a^2 - x^2} \mathrm{d} x$ and put $x = a \sin t$, with $t$ in $[-\pi/2, \pi/2]$. Here is why it gives the formula.
Why $\int 1/\sqrt{a^2 - x^2} \mathrm{d} x = \sin^{-1}(x/a) + C$
- $x = a \sin t$, so $\mathrm{d} x = a \cos t \mathrm{d} t$
Differentiate $x = a \sin t$, then write $\mathrm{d} x$ as $a \cos t \mathrm{d} t$. - $\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 t}$
Put $a \sin t$ in place of $x$. - $\sqrt{a^2 - a^2 \sin^2 t} = \sqrt{a^2 \cos^2 t} = a \cos t$
Use $1 - \sin^2 t = \cos^2 t$, and $\cos t \geq 0$ for $t$ in $[-\pi/2, \pi/2]$. - $\int 1/\sqrt{a^2 - x^2} \mathrm{d} x = \int (a \cos t)/(a \cos t) \mathrm{d} t$
Put both into the integral. - $\int 1 \mathrm{d} t = t + C$
The fraction is 1, and the integral of 1 is $t$. - *$t + C = \sin^{-1}(x/a) + C$*
$\sin t = x/a$ with $t$ on the principal branch, so $t = \sin^{-1}(x/a)$.
To use a formula, first find $a$. Take $\int 1/\sqrt{25 - x^2} \mathrm{d} x$.
Find $\int 1/\sqrt{25 - x^2} \mathrm{d} x$ from the toolkit
- $a^2 = 25$, so $a = 5$
Match the shape $1/\sqrt{a^2 - x^2}$: the number in place of $a^2$ is 25. - *$\int 1/\sqrt{25 - x^2} \mathrm{d} x = \sin^{-1}(x/5) + C$*
Put $a = 5$ into the formula $\sin^{-1}(x/a) + C$. - $d/d x (\sin^{-1}(x/5)) = 1/\sqrt{1 - x^2/25} \cdot 1/5$
Check by differentiating: the chain rule gives the extra factor $1/5$. - $1/\sqrt{1 - x^2/25} \cdot 1/5 = 1/\sqrt{25 - x^2}$
Take the 5 inside the root as 25: the integrand comes back.
Sometimes the shape is hidden. Then complete the square first. Take $\int 1/(x^2 + 4 x + 13) \mathrm{d} x$. Half of 4 is 2, and $2^2$ is 4, so split 13 as $4 + 9$:
$x^2 + 4 x + 13 = (x^2 + 4 x + 4) + 9$.
The bracket is a perfect square:
$x^2 + 4 x + 13 = (x + 2)^2 + 9$.
Now put $t = x + 2$. Its derivative is 1, so $\mathrm{d} t = \mathrm{d} x$:
$\int 1/(x^2 + 4 x + 13) \mathrm{d} x = \int 1/(t^2 + 3^2) \mathrm{d} t$.
That is the first formula, with $a = 3$.
Your turn: find $\int 1/(x^2 - 36) \mathrm{d} x$. Which formula fits, and what is $a$?
- $\int 1/(x^2+a^2) d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int 1/(x^2-a^2) d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int 1/\sqrt{a^2-x^2} d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int 1/\sqrt{9-x^2} d x$ matches which special form, and what does $a$ equal?
Check your answer
A special form, matched by shape: $\int 1/(x^2+4) \mathrm{d} x$
- Find $\int 1/(x^2+4) \mathrm{d} x$
Match the integrand to a formula in the toolkit. - $\int 1/(x^2+a^2) \mathrm{d} x = 1/a \cdot \tan^{-1}(x/a) + C$, with $a^2=4$ so $a=2$
The integrand has the shape $1/(x^2 + a^2)$, with $a^2 = 4$. - $\int 1/(x^2+4) \mathrm{d} x = 1/2 \cdot \tan^{-1}(x/2) + C$
Put $a = 2$ into the formula. There is nothing to derive. - Check by differentiating: $d/d x (1/2 \cdot \tan^{-1}(x/2)) = 1/2 \cdot 1/(1+(x/2)^2) \cdot 1/2 = 1/(4+x^2)$
Check by differentiating. The chain rule gives the extra factor $1/2$, and the result is the integrand.
- For $\int 1/(x^2+4) d x$, matching $1/(x^2+a^2)$ gives $a$ equal to \_\_\_\_\_\_.
Check your answer
- With $a=2$, $\int 1/(x^2+4) d x$ equals \_\_\_\_\_\_.
Check your answer
- Differentiating $1/2 \cdot \tan^{-1}(x/2)$ to check the answer gives \_\_\_\_\_\_.
Check your answer
Integration by parts, and when to skip it
The product rule from Chapter 5 says $d/d x (u w) = u' w + u w'$. Read it backwards and you get INTEGRATION BY PARTS.
Say you want $\int u \cdot v \mathrm{d} x$. The product rule gives a formula for it, one line at a time.
From the product rule to by parts
- $w = \int v \mathrm{d} x$, so $w' = v$
Call an integral of $v$ by the letter $w$: its derivative is $v$. - $d/d x (u w) = u' w + u w'$
The product rule from Chapter 5. - $u w = \int u' w \mathrm{d} x + \int u w' \mathrm{d} x$
Integrate both sides: the integral of a derivative is the function itself. - $u w = \int u' w \mathrm{d} x + \int u v \mathrm{d} x$
Put $v$ in place of $w'$. - $\int u v \mathrm{d} x = u w - \int u' w \mathrm{d} x$
Move $\int u' w \mathrm{d} x$ to the other side, changing its sign. - *$\int u \cdot v \mathrm{d} x = u \int v \mathrm{d} x - \int (u' \int v \mathrm{d} x) \mathrm{d} x$*
Put $\int v \mathrm{d} x$ back in place of $w$.
*The factor $u$ is the FIRST FUNCTION* and $v$ is the SECOND FUNCTION. You differentiate the first and integrate the second.
Choose them so that the new integral is easier. A factor that gets simpler when you differentiate it makes a good first function. So does a factor you cannot integrate easily, such as $\ln x$. Many teachers order the choice with the word ILATE. Its letters stand for inverse trig, log, algebraic, trig and exponential. Take as $u$ the factor that comes first. It is a rule of thumb, not part of the method.
Let us try $\int x \sin x \mathrm{d} x$, with the formula.
Find $\int x \sin x \mathrm{d} x$ by parts
- $u = x$, $v = \sin x$
Differentiating $x$ gives 1, which is simpler, so $x$ is the first function. - $\int v \mathrm{d} x = -\cos x$
The formula needs an integral of $v$, and the table gives $-\cos x$. - $\int x \sin x \mathrm{d} x = -x \cos x - \int 1 \cdot (-\cos x) \mathrm{d} x$
Put $u = x$, $u' = 1$ and $\int v \mathrm{d} x = -\cos x$ into the formula. - *$\int x \sin x \mathrm{d} x = -x \cos x + \sin x + C$*
Two minus signs make a plus, and the integral of $\cos x$ is $\sin x$. - $d/d x (-x \cos x + \sin x) = -\cos x + x \sin x + \cos x = x \sin x$
Check by differentiating with the product rule: the integrand comes back.
So one hard integral becomes an easier one.
Your turn: find $\int x \ln x \mathrm{d} x$. Which factor has to be $u$?
- Integration by parts states $\int u \cdot v d x$ equals \_\_\_\_\_\_.
Check your answer
- ILATE ranks factor types for choosing $u$. What is it actually for?
Check your answer
- Choosing which factor is $u$ and which is $v$ in by-parts matters because \_\_\_\_\_\_.
Check your answer
- For $\int \ln x \cdot x d x$, ILATE suggests \_\_\_\_\_\_.
Check your answer
By parts, checked by differentiating back: $\int x \cdot e^x \mathrm{d} x$
- Find $\int x \cdot e^x \mathrm{d} x$
A product of $x$ and $e^x$. Differentiating $x$ makes it simpler, so take $u = x$. - $u = x$, $v = e^x$
$x$ becomes 1 when you differentiate it. $e^x$ stays $e^x$ when you integrate it. - $\int v \mathrm{d} x = e^x$
The formula needs this integral of $v$. - $\int x \cdot e^x \mathrm{d} x = x \cdot e^x - \int 1 \cdot e^x \mathrm{d} x$
Put $u = x$, $u' = 1$ and $\int v \mathrm{d} x = e^x$ into the formula for by parts. - $\int x \cdot e^x \mathrm{d} x = x \cdot e^x - e^x + C$
The integral left over is in the standard table. - Check by differentiating: $d/d x (x \cdot e^x - e^x) = e^x + x \cdot e^x - e^x = x \cdot e^x$
Check by differentiating with the product rule. You get back $x \cdot e^x$.
By parts twice: $\int x^2 e^x \mathrm{d} x$
- $u = x^2$, $v = e^x$
Differentiating $x^2$ makes it simpler: take $x^2$ as the first function, same choice as $x \cdot e^x$. - $\int v \mathrm{d} x = e^x$
The formula needs this integral of $v$. - $\int x^2 e^x \mathrm{d} x = x^2 e^x - \int 2 x \cdot e^x \mathrm{d} x$
Put $u = x^2$, $u' = 2 x$ and $\int v \mathrm{d} x = e^x$ into the formula for by parts. - $\int x \cdot e^x \mathrm{d} x = x e^x - e^x + C$
A second round of by parts, on the leftover integral: the same working as $\int x \cdot e^x \mathrm{d} x$ above. - *$\int x^2 e^x \mathrm{d} x = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2 x + 2) + C$*
Put the second round’s answer back in and collect the $e^x$ terms. - $d/d x (e^x(x^2 - 2 x + 2)) = e^x(x^2 - 2 x + 2) + e^x(2 x - 2) = x^2 e^x$
Check by differentiating with the product rule: the integrand comes back.
By parts twice, then solving for the integral: $\int e^x \cos x \mathrm{d} x$
- Let $I = \int e^x \cos x \mathrm{d} x$
Name the integral $I$: it will reappear on the right side. - $u = \cos x$, $v = e^x$
Either factor can be the first function; this round takes $\cos x$. - $I = e^x \cos x + \int e^x \sin x \mathrm{d} x$
By parts: $\int v \mathrm{d} x = e^x$, and $u' = -\sin x$ turns the usual minus sign into a plus. - $u = \sin x$, $v = e^x$, for $\int e^x \sin x \mathrm{d} x$
A second round of by parts, on the new integral. - $\int e^x \sin x \mathrm{d} x = e^x \sin x - I$
The same formula; the leftover integral is $\int e^x \cos x \mathrm{d} x$ again, not a new one, so it is $I$. - $I = e^x \cos x + e^x \sin x - I$
Substitute the second round’s result back into the first line. - *$2 I = e^x(\sin x + \cos x)$, so $I = e^x(\sin x + \cos x)/2 + C$*
Move $I$ to one side and divide by 2. The $+C$ belongs to the final answer, not to $I$ on the way. - $d/d x (e^x(\sin x + \cos x)/2) = e^x \cos x$
Check by differentiating with the product rule: the integrand comes back.
- For $\int x \cdot e^x d x$, ILATE gives \_\_\_\_\_\_.
Check your answer
- With $u=x$, $d v=e^x d x$, by-parts gives $\int x \cdot e^x d x$ equals \_\_\_\_\_\_.
Check your answer
- $\int x \cdot e^x d x$ equals \_\_\_\_\_\_.
Check your answer
One formula comes straight from the product rule. For a function $f$ with a derivative, differentiate $e^x f(x)$ by the product rule:
$d/d x (e^x f(x)) = e^x f(x) + e^x f'(x)$.
*Take out $e^x$* as a common factor:
$d/d x (e^x f(x)) = e^x (f(x) + f'(x))$.
Read it backwards.
$\int e^x (f(x) + f'(x)) \mathrm{d} x = e^x f(x) + C$.
Look for $e^x$ times a function plus its own derivative, $e^x (f(x) + f'(x))$. The answer is $e^x f(x) + C$, with no by parts.
Take $f(x) = \sin x$, so $f'(x) = \cos x$. The bracket is $f(x) + f'(x)$, so the shortcut gives
$\int e^x (\sin x + \cos x) \mathrm{d} x = e^x \sin x + C$.
Check by differentiating with the product rule:
$d/d x (e^x \sin x) = e^x \sin x + e^x \cos x$, the integrand.
The order of the two terms does not matter. What matters is which term is $f$. Its derivative must be the other term.
Your turn: find $\int e^x (\sec x + \sec x \tan x) \mathrm{d} x$. Which term is $f$?
- $\int e^x (f(x)+f'(x)) d x$ equals \_\_\_\_\_\_.
Check your answer
- Why does $\int e^x (f(x)+f'(x)) d x = e^x f(x)+C$ hold?
Check your answer
- Which integral fits the $e^x (f+f')$ shortcut?
Check your answer
- For $\int e^x (1/x - 1/x^2) d x$, how does spotting the shortcut compare to working through ILATE-by-rote by-parts?
Check your answer
Spotting the shortcut inside a product: $\int e^x (1/x - 1/x^2) \mathrm{d} x$
- Find $\int e^x (1/x - 1/x^2) \mathrm{d} x$
Before trying by parts, test whether the bracket is $f(x) + f'(x)$. - $f(x) = 1/x$, so $f'(x) = -1/x^2$
The second term is the derivative of the first. - $\int e^x (f(x)+f'(x)) \mathrm{d} x = e^x f(x) + C$
This formula comes from the product rule, read backwards. - $\int e^x (1/x - 1/x^2) \mathrm{d} x = e^x/x + C$
Put $f(x) = 1/x$ into the formula. - Check by differentiating: $d/d x (e^x/x) = (e^x \cdot x - e^x)/x^2 = e^x (1/x - 1/x^2)$
Check by differentiating with the quotient rule. You get back the integrand.
- For $\int e^x (1/x - 1/x^2) d x$, matching $e^x (f+f')$ requires $f(x)$ equal to \_\_\_\_\_\_.
Check your answer
- With $f(x)=1/x$, the shortcut $\int e^x (f+f') d x = e^x f(x)+C$ gives \_\_\_\_\_\_.
Check your answer
- Differentiating $e^x/x$ to check gives $(e^x \cdot x - e^x)/x^2$, which simplifies to \_\_\_\_\_\_.
Check your answer
Here is a working for $\int e^x (x^2 + 2 x) \mathrm{d} x$. Which step is wrong?
- ILATE puts the algebraic factor first, so take $u = x^2 + 2 x$ and $v = e^x$.
- So by parts is the only way to find this integral.
- By parts once gives $(x^2 + 2 x) e^x - \int (2 x + 2) e^x \mathrm{d} x$.
Keep your answer, then look at the bracket. With $f(x) = x^2$ you get $f'(x) = 2 x$. So the bracket is $f(x) + f'(x)$, and the shortcut gives the answer in one line:
$\int e^x (x^2 + 2 x) \mathrm{d} x = x^2 e^x + C$.
Check by differentiating with the product rule:
$d/d x (x^2 e^x) = 2 x e^x + x^2 e^x$, the integrand.
The second step is wrong. ILATE only orders the choice of $u$. It does not say that by parts is the best way. By parts does reach $x^2 e^x + C$ too, but only after a second round. Each round is one more chance to drop a sign.
So look for a pattern before you reach for by parts.
I saw $\int e^x (\sin x + \cos x) \mathrm{d} x$ and went straight to by parts, taking $u = \sin x$ and $v = e^x$ because ILATE puts the trigonometric factor ahead of the exponential.
Check the $e^x (f(x)+f'(x))$ shortcut FIRST, before reaching for by parts at all. Here $f(x) = \sin x$ gives $f'(x) = \cos x$, and the bracket $\sin x + \cos x$ is exactly $f(x)+f'(x)$.
$\int e^x (f(x)+f'(x)) \mathrm{d} x = e^x f(x) + C$ gives $e^x \sin x + C$ in one line. By parts reaches the same answer, but only after two full rounds, each one a fresh chance to drop a sign.
- A student sees $\int e^x (x^2+2x) d x$, checks ILATE, sets $u=x^2+2x$, and works through two rounds of by parts. What did the student miss?
Check your answer
- ILATE is best understood as \_\_\_\_\_\_.
Check your answer
- For $\int e^x (x^2+2x) d x$, compare the shortcut answer to the ILATE-by-rote path.
Check your answer
Partial fractions before integrating
In Class 7 you added fractions. One half plus one third is five sixths. You made one denominator, 6, out of the two denominators 2 and 3. Now run it the other way. Start from five sixths. Since 6 is 2 times 3, you can split five sixths back into a half and a third.
So a fraction whose denominator is a product can be split. You get simpler fractions, one for each factor.
The same works for functions. A RATIONAL FUNCTION is one polynomial divided by another. It is PROPER when the degree of the numerator is less than the degree of the denominator. Say the denominator has two different linear factors. Then a proper rational function splits into PARTIAL FRACTIONS.
$(p x + q)/((x - a)(x - b)) = A/(x - a) + B/(x - b)$, with $a \neq b$.
Each piece integrates to a log: $\int A/(x - a) \mathrm{d} x = A \ln|x - a| + C$.
Let us split $(2 x + 1)/((x - 1)(x + 2))$, then integrate it.
Split $(2 x + 1)/((x - 1)(x + 2))$ and integrate
- $(2 x + 1)/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2)$
Write one fraction for each factor of the denominator. - $2 x + 1 = A(x + 2) + B(x - 1)$
Multiply both sides by $(x - 1)(x + 2)$, which clears every denominator. - At $x = 1$: $3 = 3 A$, so $A = 1$
Put $x = 1$: it makes $x - 1$ zero, so the $B$ term drops out. - At $x = -2$: $-3 = -3 B$, so $B = 1$
Put $x = -2$: it makes $x + 2$ zero, so the $A$ term drops out. - *$\int (2 x + 1)/((x - 1)(x + 2)) \mathrm{d} x = \ln|x - 1| + \ln|x + 2| + C$*
Each piece integrates to a log. - At $x = 0$: $-1 + 1/2 = -1/2$, and $1/(-2) = -1/2$
Check the split at $x = 0$: the two pieces and the fraction agree.
Two more splits are needed for Exercise 7.5. A repeated linear factor needs a term for each power of $(x - a)$:
$(p x + q)/((x - a)^2(x - b)) = A/(x - a) + B/(x - a)^2 + D/(x - b)$.
An irreducible quadratic factor, one with no real linear factors, needs a linear numerator over it, not a bare constant:
$(p x + q)/((x - a)(x^2 + b x + c)) = A/(x - a) + (B x + D)/(x^2 + b x + c)$.
A repeated factor: $\int (6 x + 2)/((x - 1)^2(x + 3)) \mathrm{d} x$
- $(6 x + 2)/((x - 1)^2(x + 3)) = A/(x - 1) + B/(x - 1)^2 + D/(x + 3)$
One factor is repeated: it needs both a plain and a squared term. - $6 x + 2 = A(x - 1)(x + 3) + B(x + 3) + D(x - 1)^2$
Multiply both sides by $(x - 1)^2(x + 3)$, which clears every denominator. - At $x = 1$: $8 = 4 B$, so $B = 2$
Substituting the repeated root kills the $A$ and $D$ terms, leaving $B$ directly. - At $x = -3$: $-16 = 16 D$, so $D = -1$
Substituting the other root kills the $A$ and $B$ terms, leaving $D$ directly. - Comparing $x^2$-coefficients: $A + D = 0$, so $A = 1$
The left side has no $x^2$-term, so $A$ and $D$’s $x^2$-contributions must cancel. - *$\int (6 x + 2)/((x - 1)^2(x + 3)) \mathrm{d} x = \ln|x - 1| - 2/(x - 1) - \ln|x + 3| + C$*
The plain-factor terms use the log rule; the squared term uses the power rule with $n = -2$. - At $x = 0$: $1/(0-1) + 2/(0-1)^2 - 1/(0+3) = -1 + 2 - 1/3 = 2/3$, and the fraction at $x = 0$ is $2/((-1)^2 \cdot 3) = 2/3$
Check by putting $x = 0$ into the split and into the original fraction: both give the same value.
A quadratic factor: $\int (3 x^2 - 5 x + 8)/((x - 3)(x^2 + 2 x + 5)) \mathrm{d} x$
- $(3 x^2 - 5 x + 8)/((x - 3)(x^2 + 2 x + 5)) = A/(x - 3) + (B x + D)/(x^2 + 2 x + 5)$
An irreducible quadratic factor needs a linear numerator over it. - $3 x^2 - 5 x + 8 = A(x^2 + 2 x + 5) + (B x + D)(x - 3)$
Multiply both sides by $(x - 3)(x^2 + 2 x + 5)$. - At $x = 3$: $20 = 20 A$, so $A = 1$
Substituting the linear factor’s root kills the quadratic term, leaving $A$ directly. - Comparing $x^2$-coefficients: $3 = A + B$, so $B = 2$
The $x^2$-terms on both sides must match. - Comparing constant terms: $8 = 5 A - 3 D$, so $D = -1$
The constant terms on both sides must match, once $A$ is known. - $\int [1/(x - 3) + (2 x - 1)/(x^2 + 2 x + 5)] \mathrm{d} x$
Split the second term: $2 x - 1 = (2 x + 2) - 3$, the first part the derivative of $x^2 + 2 x + 5$. - *$= \ln|x - 3| + \ln(x^2 + 2 x + 5) - 3/2 \cdot \tan^{-1}((x + 1)/2) + C$*
The log term comes from $u = x^2 + 2 x + 5$; the last from $(x + 1)^2 + 4$, the $1/(x^2+a^2)$ special form with $a = 2$. - At $x = 0$: $1/(0-3) + (2 \cdot 0 - 1)/(0 + 0 + 5) = -1/3 - 1/5 = -8/15$, and the fraction at $x = 0$ is $8/((-3) \cdot 5) = -8/15$
Check by putting $x = 0$ into the split and into the original fraction: both give the same value.
If the degree of the numerator is not less, the function is IMPROPER. Divide first. What is left over is proper, and you split that.
Your turn: split $1/((x - 1)(x + 1))$ into partial fractions. Do not integrate yet.
Dividing first: $\int x^3/(x + 2) \mathrm{d} x$
- $x^3/(x + 2)$
Improper: the top’s degree, 3, is not less than the bottom’s, 1. Divide first. - $x^3 = (x + 2)(x^2 - 2 x + 4) - 8$
Long division of $x^3$ by $x + 2$: the quotient is $x^2 - 2 x + 4$, the remainder $-8$. - $x^3/(x + 2) = x^2 - 2 x + 4 - 8/(x + 2)$
Divide both sides by $x + 2$. The polynomial part splits off; what is left, $-8/(x+2)$, is proper. - $\int x^3/(x + 2) \mathrm{d} x = \int (x^2 - 2 x + 4) \mathrm{d} x - 8 \int 1/(x + 2) \mathrm{d} x$
Integrate the polynomial part and the proper part term by term. - *$\int x^3/(x + 2) \mathrm{d} x = x^3/3 - x^2 + 4 x - 8 \ln|x + 2| + C$*
Each piece is a standard integral; the log rule gives the last one. - $(x + 2)(x^2 - 2 x + 4) - 8 = x^3$
Check the division: multiply the quotient back by $x + 2$ and subtract $8$; the cubic returns.
- For distinct linear factors, $P(x)/((x-a)(x-b))$ splits into \_\_\_\_\_\_.
Check your answer
- After clearing denominators to get $P(x) = A(x-b)+B(x-a)$, why does substituting $x=a$ quickly give $A$?
Check your answer
- Once $P(x)/((x-1)(x+2)) = 2/(x-1) + 1/(x+2)$ is found, integrating gives \_\_\_\_\_\_.
Check your answer
- A rational function has numerator degree 3 and denominator degree 2. What must happen before partial fractions apply?
Check your answer
Partial fractions, term by term: $\int (3x+3)/((x-1)(x+2)) \mathrm{d} x$
- Find $\int (3x+3)/((x-1)(x+2)) \mathrm{d} x$
The numerator has degree 1 and the denominator degree 2, so the fraction is proper. The two factors are different. - $(3x+3)/((x-1)(x+2)) = A/(x-1) + B/(x+2)$
Write one fraction for each factor. - $3x+3 = A(x+2) + B(x-1)$
Multiply both sides by $(x - 1)(x + 2)$. - At $x=1$: $6=3A$, so $A=2$. At $x=-2$: $-3=-3B$, so $B=1$
Each root makes one bracket zero. So each leaves one unknown. - $\int [2/(x-1) + 1/(x+2)] \mathrm{d} x = 2 \ln|x-1| + \ln|x+2| + C$
Each piece is a log formula from the standard table. - Check: $2/(x-1)+1/(x+2) = (2(x+2)+(x-1))/((x-1)(x+2)) = (3x+3)/((x-1)(x+2))$
Check by adding the two fractions back. You get the integrand.
- Substituting $x=1$ into $3x+3=A(x+2)+B(x-1)$ gives $6=3A$, so $A$ equals \_\_\_\_\_\_.
Check your answer
- Substituting $x=-2$ into $3x+3=A(x+2)+B(x-1)$ gives $-3=-3B$, so $B$ equals \_\_\_\_\_\_.
Check your answer
- With $A=2$, $B=1$, $\int (3x+3)/((x-1)(x+2)) d x$ equals \_\_\_\_\_\_.
Check your answer
Definite integrals by substitution
So far a definite integral took two steps: find the integral, then put in the limits. When the integral needs a substitution, you have a choice. The long way: substitute, integrate in $t$, put $x$ back, and then put in the limits in $x$. That is the substitution section’s integral, finished with the Fundamental Theorem. The short way: change the limits to $t$ at the moment you substitute, and never come back to $x$. The board’s answers use the short way. Here it is.
Put $t = g(x)$. Then $\mathrm{d} t = g'(x) \mathrm{d} x$, as in the substitution section. Change each limit with $t$ too:
$x = a$ gives $t = g(a)$.
$x = b$ gives $t = g(b)$.
*Put in $t$*, $\mathrm{d} t$ and the new limits:
$\int_a^b f(g(x)) \cdot g'(x) \mathrm{d} x = \int_{g(a)}^{g(b)} f(t) \mathrm{d} t$.
After that you work in $t$ alone. There is no need to go back to $x$.
Let us find $\int_0^2 2 x e^{x^2} \mathrm{d} x$ the short way.
Find $\int_0^2 2 x e^{x^2} \mathrm{d} x$ with new limits
- $t = x^2$, so $\mathrm{d} t = 2 x \mathrm{d} x$
The derivative of $x^2$ is $2 x$, and $2 x \mathrm{d} x$ appears in the integral. - $x = 0$ gives $t = 0$, and $x = 2$ gives $t = 4$
Change each limit with $t = x^2$. - $\int_0^2 2 x e^{x^2} \mathrm{d} x = \int_0^4 e^t \mathrm{d} t$
Put in $t$, $\mathrm{d} t$ and the new limits: now everything is in $t$. - $\int_0^4 e^t \mathrm{d} t = [e^t]_0^4 = e^4 - e^0$
The integral of $e^t$ is $e^t$, at 4 minus at 0. - *$\int_0^2 2 x e^{x^2} \mathrm{d} x = e^4 - 1$*
Any number to the power 0 is 1.
If you keep the old limits 0 and 2 with $t$, you get $e^2 - 1$. That is wrong, because $t$ runs from 0 to 4. Check with $F(x) = e^{x^2}$ in $x$. It gives $e^4 - e^0$, the same $e^4 - 1$.
Your turn: find $\int_1^2 2 x (x^2 - 1)^3 \mathrm{d} x$ with $t = x^2 - 1$. What are the new limits?
- Substituting $t=g(x)$ in a definite integral $\int_a^b f(g(x)) \cdot g'(x) d x$ changes \_\_\_\_\_\_.
Check your answer
- For $\int_a^b f(g(x)) \cdot g'(x) d x$ with $t=g(x)$, the new limits are \_\_\_\_\_\_.
Check your answer
- Once the limits are changed to $t=g(a)$ and $t=g(b)$, what happens at the end of a definite-integral substitution?
Check your answer
A definite integral by substitution: $\int_0^1 2x \sqrt{x^2+1} \mathrm{d} x$
- Find $\int_0^1 2x \sqrt{x^2+1} \mathrm{d} x$
Substitute, and change the limits along with the variable. - $t = x^2+1$, $\mathrm{d} t = 2x \mathrm{d} x$; when $x=0$, $t=1$; when $x=1$, $t=2$
Work out $t$ at each limit. - $\int_0^1 2x \sqrt{x^2+1} \mathrm{d} x = \int_1^2 \sqrt{t} \mathrm{d} t$
Now everything is in $t$, the limits too. - $\int_1^2 \sqrt{t} \mathrm{d} t = [2/3 \cdot t^{3/2}]_1^2 = 2/3 \cdot (2 \sqrt{2} - 1)$
Use the power rule in $t$. Then subtract its value at $t = 1$ from its value at $t = 2$. - Check against the earlier integral $2/3 \cdot (x^2+1)^{3/2}$: at $x=1$ it gives $2/3 \cdot 2 \sqrt{2}$, at $x=0$ it gives $2/3$
Check with the integral in $x$ from the first worked example. The difference is the same.
- With $t=x^2+1$, the limits for $\int_0^1 2x \sqrt{x^2+1} d x$ become \_\_\_\_\_\_.
Check your answer
- $\int_1^2 \sqrt{t} d t = [2/3 \cdot t^{3/2}]_1^2$ equals \_\_\_\_\_\_.
Check your answer
- Why is $2/3 \cdot (2 \sqrt{2} - 1)$ the final answer, with no further step needed?
Check your answer
Properties of definite integrals
A few properties make some definite integrals easy, often without finding an integral at all. In each one, let $f$ be continuous on the interval.
- The letter does not matter. $\int_a^b f(x) \mathrm{d} x = \int_a^b f(t) \mathrm{d} t$.
- Turning the limits round changes the sign. $\int_a^b f(x) \mathrm{d} x = -\int_b^a f(x) \mathrm{d} x$, so $\int_a^a f(x) \mathrm{d} x = 0$.
- You can split the interval at any $c$. $\int_a^b f(x) \mathrm{d} x = \int_a^c f(x) \mathrm{d} x + \int_c^b f(x) \mathrm{d} x$.
- Putting $a + b - x$ for $x$ keeps the value. $\int_a^b f(x) \mathrm{d} x = \int_a^b f(a + b - x) \mathrm{d} x$.
- On $[0, a]$ the same rule reads $\int_0^a f(x) \mathrm{d} x = \int_0^a f(a - x) \mathrm{d} x$.
- On $[-a, a]$, an EVEN function, with $f(-x) = f(x)$, gives $2 \int_0^a f(x) \mathrm{d} x$. An ODD function, with $f(-x) = -f(x)$, gives 0.
- On $[0, 2 a]$, $\int_0^{2 a} f(x) \mathrm{d} x = 2 \int_0^a f(x) \mathrm{d} x$ if $f(2 a - x) = f(x)$, and $0$ if $f(2 a - x) = -f(x)$. To see it, split at $a$ and put $x = 2 a - t$ in the second piece.
Check the even and odd rule with numbers. $x^2$ is even. From 0 to 2,
$\int_0^2 x^2 \mathrm{d} x = 8/3$.
The piece on $[-2, 0]$ is its mirror image, also $8/3$, so the whole is twice as much:
$\int_{-2}^2 x^2 \mathrm{d} x = 16/3$.
*Now $x^3$*, which is odd. Its piece on $[-2, 0]$ is $-4$ and its piece on $[0, 2]$ is 4, so they cancel:
$\int_{-2}^2 x^3 \mathrm{d} x = 0$.
The reflection rule helps when $f(a + b - x)$ is simpler, or pairs well with $f(x)$. On $[0, \pi/2]$, put $\pi/2 - x$ in place of $x$:
$\sin(\pi/2 - x) = \cos x$.
So the rule turns $\sin x$ into $\cos x$. The two have the same integral on $[0, \pi/2]$.
$\int_0^{\pi/2} \sin x \mathrm{d} x = \int_0^{\pi/2} \cos x \mathrm{d} x$.
Each one is 1. An integral of $\sin x$ is $-\cos x$, which is 0 at $\pi/2$ and $-1$ at 0, and $0 - (-1)$ is 1.
The reflection move, added to itself: $\int_0^{\pi/2} \sin x/(\sin x + \cos x) \mathrm{d} x$
- Let $I = \int_0^{\pi/2} \sin x/(\sin x + \cos x) \mathrm{d} x$
Name the integral $I$, so it can be combined with its own reflection. - $I = \int_0^{\pi/2} \cos x/(\cos x + \sin x) \mathrm{d} x$
Replace $x$ by $\pi/2 - x$: $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$, and the limits stay $0$ to $\pi/2$. - $I + I = \int_0^{\pi/2} (\sin x + \cos x)/(\sin x + \cos x) \mathrm{d} x$
Add the two equal forms of $I$: the denominators match, so the fractions combine into one. - *$2 I = \int_0^{\pi/2} 1 \mathrm{d} x = \pi/2$*
The integrand is $1$ everywhere on $[0, \pi/2]$, so the integral is just the length of the interval. - $I = \pi/4$
Divide by $2$. - $f(x) + f(\pi/2 - x) = 1$ for every $x$ in $[0, \pi/2]$
Step 2 gives this for every $x$, not only at one point, so the average height over the interval is $1/2$: the area is $(1/2)(\pi/2) = \pi/4$, matching.
One more is true, though NCERT does not list it. If $f$ repeats every $p$, then $\int_0^{n p} f(x) \mathrm{d} x = n \int_0^p f(x) \mathrm{d} x$.
Your turn: find $\int_{-2}^2 x^2 \sin x \mathrm{d} x$ without integrating. Is $x^2 \sin x$ even or odd?
- The reversal property states $\int_a^b f(x) d x$ equals \_\_\_\_\_\_.
Check your answer
- The reflection property, $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$, is useful when \_\_\_\_\_\_.
Check your answer
- For $f$ ODD on $[-a,a]$, $\int_{-a}^a f(x) d x$ equals \_\_\_\_\_\_.
Check your answer
- If $f$ has period $p$, the periodicity property gives $\int_0^{n p} f(x) d x$ as \_\_\_\_\_\_.
Check your answer
Every method is the same question, asked differently
The question at the start asked if you can go back, from the rate to the quantity.
- INTEGRAL: $\int f(x) \mathrm{d} x = F(x) + C$. A flow of 20 litres a minute gives $20 t + C$ litres.
- SUBSTITUTION: the chain rule read backwards. $\int 2 x \cos(x^2) \mathrm{d} x = \sin(x^2) + C$.
- BY PARTS: the product rule read backwards. $\int x \sin x \mathrm{d} x = -x \cos x + \sin x + C$.
- PARTIAL FRACTIONS: adding fractions undone. $(2 x + 1)/((x - 1)(x + 2)) = 1/(x - 1) + 1/(x + 2)$.
- DEFINITE INTEGRAL: a limit of totals of thin strips. Under $x^2/4 + 1$ on $[0, 4]$ the totals climb to $28/3$.
- FUNDAMENTAL THEOREM: $\int_a^b f(x) \mathrm{d} x = F(b) - F(a)$. So $\int_1^2 (3 x^2 - 2 x) \mathrm{d} x = 4 - 0 = 4$.
Here is the answer. Yes, you can go back, but you get a family of functions. There is one for every starting amount, and that is the $+ C$. Each method is a rule of differentiation read backwards. Substitution is the chain rule, by parts is the product rule, and partial fractions is adding fractions undone. The total change between two moments is the area under the graph of the rate. It is the limit of totals of thin strips. The fundamental theorem says it equals $F(b) - F(a)$, so the $C$ cancels.
- Every technique in this chapter answers one central question. What is it?
Check your answer
- When no technique matches an integrand directly, the next step is to \_\_\_\_\_\_.
Check your answer
- A definite integral adds one more question on top of finding an integral. What is it?
Check your answer
- A student proposes $\int 3x^2 d x = x^3 + C$. What is the correct way to check whether this is right?
Check your answer
- Which of these correctly completes the standard integrals table for $\int \sin x d x$?
Check your answer
- For $\int x \cos(x^2) d x$, take $t = x^2$, so $d t = 2x d x$. But the integrand only has $x d x$, not $2x d x$. What should be done?
Check your answer
- Integration by parts subtracts a second integral: $\int u \cdot v d x = u \int v d x - \int (d/(d x) (u) \cdot \int v d x) d x$. For $\int x \cdot e^x d x$ with $u = x$ and $v = e^x$, a student ends with $x \cdot e^x + e^x + C$ instead of $x \cdot e^x - e^x + C$. What went wrong?
Check your answer
- Using the second fundamental theorem on $\int_0^3 2x d x$ with integral $F(x) = x^2$, a student writes the answer as $F(0) - F(3) = 0 - 9 = -9$. What went wrong?
Check your answer
Where you will meet this
Each one below is the same idea you already use, just named differently.
- A startup’s monthly profit. The founder tracks how fast profit is growing each month.
The maths: A constant growth rate, added onto a starting balance, is exactly an integral plus its constant of integration.
A startup’s profit grows ₹$500$ a month, starting from ₹$2000$ already banked. After $6$ months the total is ₹$5000$.
- A recipe’s unit conversion. A cook scales a recipe from cups to grams.
The maths: Swapping one unit for another inside a sum is the same move substitution makes inside an integral.
$1$ cup of rice is $200$ g, so a recipe needing $600$ g of rice calls for $600/200=3$ cups.
- A tailor’s combined bill. A tailor prices cloth and labour as one combined total.
The maths: Splitting one combined total back into its separate parts is the same move partial fractions makes on a single fraction.
A combined bill of ₹$700$ splits into a cloth part of ₹$450$ and a labour part of ₹$250$. Together $450+250=700$.
- A marathon runner’s pace. A runner holds a steady pace for part of a race.
The maths: A steady pace held over many short strips of time adds up to a total distance. That total, taken to its limit, is exactly what a definite integral stands for.
A steady pace of $10$ km/h held for $0.5$ hour adds $10 \times 0.5=5$ km to the total distance.
- A shop’s hourly count of people coming in and going out. A shop counts people entering and leaving across one hour.
The maths: A net change can look small while the actual number of people coming in and going out is much larger. A signed definite integral hides that same gap from a plain area count.
$40$ customers enter and $25$ leave in an hour. The net change is $40-25=15$, even though $65$ people actually moved through the door.
Your turn: A café’s cash box starts the day with ₹$500$ and takes in ₹$200$ every hour it is open. How much is in the cash box after $5$ hours?
Where integrals go next
First-year engineering mathematics extends this chapter’s integration to functions of two or more variables, as a multiple integral taken over a region instead of an interval. It also meets the Laplace transform, which integrates a function against $e^{-s t}$ to turn a differential equation into an algebraic one. Both methods build on the single-variable integral this chapter sets up.
Practice set: Exercise 7.1
Exercise 7.1 practises the formulas of the standard table. Some questions need a little algebra first, such as opening a bracket or splitting a fraction. Check each answer by differentiating it.
- practice Find $\int \sin 5x \mathrm{d} x$.
- practice Find $\int \cos 7x \mathrm{d} x$.
- practice Find $\int e^{4x} \mathrm{d} x$.
- practice Find $\int x^6 \mathrm{d} x$.
- practice Which of these equals $\int 3/x \mathrm{d} x$?
- practice Find $\int \sec^2 3x \mathrm{d} x$.
- practice Find $\int \text{cosec}^2 x \mathrm{d} x$.
- practice Find $\int \sec x \tan x \mathrm{d} x$.
- practice Find $\int \text{cosec} x \cot x \mathrm{d} x$.
- practice Find $\int 4 e^x \mathrm{d} x$.
- practice Find $\int (3 \sin x - 2 \cos x) \mathrm{d} x$.
- practice Find $\int (2x - 5)^2 \mathrm{d} x$.
- practice Find $\int x^2 (x^3 + 1) \mathrm{d} x$.
- practice Find $\int (x^2 + 3x + 2)/x \mathrm{d} x$.
- practice Find $\int (x^3 - 2x + 1)/x^2 \mathrm{d} x$.
- practice Find $\int \text{cosec} x (\text{cosec} x - \cot x) \mathrm{d} x$.
- practice Find $\int (x+1)(x+2) \mathrm{d} x$.
- practice Find $\int (\sqrt{x} + 1/\sqrt{x})^2 \mathrm{d} x$.
- practice Find $\int (x^3 - 3x^2 + 5)/\sqrt{x} \mathrm{d} x$.
- practice Find $\int (2x+3)(x-1) \mathrm{d} x$.
- practice Which of these equals $\int (x^4 + 1/x^4) \mathrm{d} x$?
- practice If $d y/d x = 4x^3 - 3/x^4$ and $y = 2$ when $x = 1$, find $y$ as a function of $x$.
- practice Find $\int 3/(x \sqrt{x^2 - 1}) \mathrm{d} x$, for $x > 1$.
Answers
- $-(1/5) \cos 5x + C$
- $(1/7) \sin 7x + C$
- $(1/4) e^{4x} + C$
- $x^7/7 + C$
- (D) $3 \ln|x| + C$.
- $(1/3) \tan 3x + C$
- $-\cot x + C$
- $\sec x + C$
- $-\text{cosec} x + C$
- $4 e^x + C$
- $-3 \cos x - 2 \sin x + C$
- $(4/3) x^3 - 10 x^2 + 25 x + C$
- $x^6/6 + x^3/3 + C$
- $x^2/2 + 3x + 2 \ln|x| + C$
- $x^2/2 - 2 \ln|x| - 1/x + C$
- $-\cot x + \text{cosec} x + C$
- $x^3/3 + (3/2) x^2 + 2x + C$
- $x^2/2 + 2x + \ln|x| + C$
- $(2/7) x^{7/2} - (6/5) x^{5/2} + 10 x^{1/2} + C$
- $(2/3) x^3 + (1/2) x^2 - 3x + C$
- (B) $x^5/5 - 1/(3x^3) + C$.
- $y = x^4 + 1/x^3$
- $3 \sec^{-1} x + C$
Practice set: Exercise 7.2
Exercise 7.2 practises substitution. In each question, look for a function whose derivative also appears in the integrand.
- practice Find $\int 2x (x^2+5)^6 \mathrm{d} x$.
- practice Find $\int 4x^3 (x^4+1)^4 \mathrm{d} x$.
- practice Find $\int x/(x^2+9) \mathrm{d} x$.
- practice Find $\int x^2/(x^3+4) \mathrm{d} x$.
- practice Find $\int (\ln x)^3/x \mathrm{d} x$.
- practice Which of these equals $\int 1/(x \ln x) \mathrm{d} x$?
- practice Find $\int e^x/(1+e^x) \mathrm{d} x$.
- practice Find $\int e^x \sin(e^x) \mathrm{d} x$.
- practice Find $\int e^x \cos(e^x) \mathrm{d} x$.
- practice Find $\int \sin x \cos^5 x \mathrm{d} x$.
- practice Find $\int \cos x \sin^3 x \mathrm{d} x$.
- practice Find $\int \sec^2 x \tan^4 x \mathrm{d} x$.
- practice Find $\int \text{cosec}^2 x \cot^3 x \mathrm{d} x$.
- practice Find $\int \sec x \tan x \sqrt{\sec x} \mathrm{d} x$.
- practice Find $\int \text{cosec} x \cot x \sqrt{\text{cosec} x} \mathrm{d} x$.
- practice Find $\int x e^{x^2} \mathrm{d} x$.
- practice Find $\int x^3 e^{x^4} \mathrm{d} x$.
- practice Find $\int x \sin(x^2) \mathrm{d} x$.
- practice Find $\int x \cos(x^2) \mathrm{d} x$.
- practice Find $\int (2x+1)/(x^2+x+3) \mathrm{d} x$.
- practice Find $\int (3x^2+2x)/(x^3+x^2+1) \mathrm{d} x$.
- practice Find $\int x \sqrt{x^2+4} \mathrm{d} x$.
- practice Find $\int x \sqrt{1-x^2} \mathrm{d} x$.
- practice Find $\int x^3 \sqrt{x^4+1} \mathrm{d} x$.
- practice Find $\int (x+1)/\sqrt{x^2+2x+5} \mathrm{d} x$.
- practice Find $\int (2x-1)/\sqrt{x^2-x+1} \mathrm{d} x$.
- practice Find $\int x/(1+x^4) \mathrm{d} x$.
- practice Find $\int e^{2x}/(1+e^{4x}) \mathrm{d} x$.
- practice Find $\int 1/(x (1+(\ln x)^2)) \mathrm{d} x$.
- practice Find $\int \sin(\sqrt{x})/\sqrt{x} \mathrm{d} x$.
- practice Find $\int \cos(\sqrt{x})/\sqrt{x} \mathrm{d} x$.
- practice Find $\int x^2/(x^3+1)^2 \mathrm{d} x$.
- practice Find $\int (x-1)/(x^2-2x+5) \mathrm{d} x$.
- practice Find $\int x^3/(x^2+1) \mathrm{d} x$.
- practice Find $\int x^3 \sqrt{x^2+1} \mathrm{d} x$.
- practice Find $\int \sec^2 x/(\tan^2 x+4 \tan x+5) \mathrm{d} x$.
- practice Find $\int 1/(\sqrt{x} (1+x)) \mathrm{d} x$.
- practice Find $\int 1/(\sqrt{x} (1+\sqrt{x})^2) \mathrm{d} x$.
- practice Which of these equals $\int 2x/(1+x^2)^2 \mathrm{d} x$?
- practice Find $\int x^3/\sqrt{1-x^2} \mathrm{d} x$.
- practice Find $\int \tan x \mathrm{d} x$.
- practice Find $\int \cot x \mathrm{d} x$.
- practice Find $\int \sec x \mathrm{d} x$.
- practice Find $\int \text{cosec} x \mathrm{d} x$.
- practice Find $\int 2^x \mathrm{d} x$.
Answers
- $(x^2+5)^7/7 + C$
- $(x^4+1)^5/5 + C$
- $(1/2) \ln|x^2+9| + C$
- $(1/3) \ln|x^3+4| + C$
- $(\ln x)^4/4 + C$
- (A) $\ln|\ln x| + C$.
- $\ln(1+e^x) + C$
- $-\cos(e^x) + C$
- $\sin(e^x) + C$
- $-\cos^6 x/6 + C$
- $\sin^4 x/4 + C$
- $\tan^5 x/5 + C$
- $-\cot^4 x/4 + C$
- $(2/3) (\sec x)^{3/2} + C$
- $-(2/3) (\text{cosec} x)^{3/2} + C$
- $(1/2) e^{x^2} + C$
- $(1/4) e^{x^4} + C$
- $-(1/2) \cos(x^2) + C$
- $(1/2) \sin(x^2) + C$
- $\ln|x^2+x+3| + C$
- $\ln|x^3+x^2+1| + C$
- $(1/3) (x^2+4)^{3/2} + C$
- $-(1/3) (1-x^2)^{3/2} + C$
- $(1/6) (x^4+1)^{3/2} + C$
- $\sqrt{x^2+2x+5} + C$
- $2 \sqrt{x^2-x+1} + C$
- $(1/2) \tan^{-1}(x^2) + C$
- $(1/2) \tan^{-1}(e^{2x}) + C$
- $\tan^{-1}(\ln x) + C$
- $-2 \cos(\sqrt{x}) + C$
- $2 \sin(\sqrt{x}) + C$
- $-1/(3(x^3+1)) + C$
- $(1/2) \ln|x^2-2x+5| + C$
- $x^2/2 - (1/2) \ln|x^2+1| + C$
- $(1/5) (x^2+1)^{5/2} - (1/3) (x^2+1)^{3/2} + C$
- $\tan^{-1}(\tan x + 2) + C$
- $2 \tan^{-1}(\sqrt{x}) + C$
- $-2/(1+\sqrt{x}) + C$
- (A) $-1/(1+x^2) + C$.
- $-\sqrt{1-x^2} + (1/3) (1-x^2)^{3/2} + C$
- $\ln|\sec x| + C$
- $\ln|\sin x| + C$
- $\ln|\sec x + \tan x| + C$
- $\ln|\text{cosec} x - \cot x| + C$
- $2^x/(\ln 2) + C$
Practice set: Exercise 7.3
Exercise 7.3 practises integrals of trigonometric functions. First rewrite the integrand with an identity, such as $\sin^2 x = (1 - \cos 2 x)/2$. Then a formula from the table applies.
- practice Find $\int \sin^2 x \mathrm{d} x$.
- practice Find $\int \cos^2 x \mathrm{d} x$.
- practice Find $\int \sin^2 (3x) \mathrm{d} x$.
- practice Find $\int \tan^2 x \mathrm{d} x$.
- practice Find $\int \cot^2 x \mathrm{d} x$.
- practice Find $\int \sec^2 x \tan x \mathrm{d} x$.
- practice Find $\int \text{cosec}^2 x \cot x \mathrm{d} x$.
- practice Find $\int \sin 4x \cos 2x \mathrm{d} x$.
- practice Find $\int \cos 3x \cos x \mathrm{d} x$.
- practice Find $\int \sin 5x \sin x \mathrm{d} x$.
- practice Find $\int \sin^3 x \mathrm{d} x$.
- practice Find $\int \cos^3 x \mathrm{d} x$.
- practice Find $\int \sin^4 x \mathrm{d} x$.
- practice Find $\int \cos^4 x \mathrm{d} x$.
- practice Find $\int \sin^2 x \cos^2 x \mathrm{d} x$.
- practice Find $\int (\sin x + \cos x)^2 \mathrm{d} x$.
- practice Find $\int \sec^4 x \mathrm{d} x$.
- practice Find $\int \text{cosec}^4 x \mathrm{d} x$.
- practice Find $\int (1-\cos x)/(1+\cos x) \mathrm{d} x$.
- practice Find $\int \sin^3 x \cos^2 x \mathrm{d} x$.
- practice Find $\int \cos^3 x \sin^2 x \mathrm{d} x$.
- practice Find $\int \sin x \sin 2x \sin 3x \mathrm{d} x$.
- practice To evaluate $\int \sin x \cos 5x \mathrm{d} x$, which identity is applied first?
- practice $\int 2 \sin^2 x \mathrm{d} x$ equals
Answers
- $x/2 - \sin(2x)/4 + C$
- $x/2 + \sin(2x)/4 + C$
- $x/2 - \sin(6x)/12 + C$
- $\tan x - x + C$
- $-\cot x - x + C$
- $\tan^2 x/2 + C$
- $-\cot^2 x/2 + C$
- $-\cos(6x)/12 - \cos(2x)/4 + C$
- $\sin(2x)/4 + \sin(4x)/8 + C$
- $\sin(4x)/8 - \sin(6x)/12 + C$
- $-\cos x + \cos^3 x/3 + C$
- $\sin x - \sin^3 x/3 + C$
- $3x/8 - \sin(2x)/4 + \sin(4x)/32 + C$
- $3x/8 + \sin(2x)/4 + \sin(4x)/32 + C$
- $x/8 - \sin(4x)/32 + C$
- $x - \cos(2x)/2 + C$
- $\tan x + \tan^3 x/3 + C$
- $-\cot x - \cot^3 x/3 + C$
- $2 \tan(x/2) - x + C$
- $-\cos^3 x/3 + \cos^5 x/5 + C$
- $\sin^3 x/3 - \sin^5 x/5 + C$
- $-\cos(4x)/16 + \cos(6x)/24 - \cos(2x)/8 + C$
- (A) $\sin A \cos B = (1/2) [\sin(A+B)+\sin(A-B)]$.
- (A) $x - \sin(2x)/2 + C$.
Practice set: Exercise 7.4, part 1
These questions are the first part of Exercise 7.4. They use the formulas for $1/(x^2 + a^2)$ and $1/(x^2 - a^2)$. Some need you to complete the square first.
- practice Find $\int 1/(x^2+9) \mathrm{d} x$.
- practice Find $\int 1/(x^2+16) \mathrm{d} x$.
- practice Find $\int 1/(x^2+1) \mathrm{d} x$.
- practice Find $\int 1/(x^2-9) \mathrm{d} x$.
- practice Find $\int 1/(x^2-25) \mathrm{d} x$.
- practice Find $\int 1/(x^2-1) \mathrm{d} x$.
- practice Find $\int 1/(x^2+36) \mathrm{d} x$.
- practice Find $\int 1/(x^2-16) \mathrm{d} x$.
- practice Find $\int 1/(x^2+49) \mathrm{d} x$.
- practice Find $\int 1/(4x^2+9) \mathrm{d} x$.
- practice Find $\int 1/(9x^2-4) \mathrm{d} x$.
- practice Find $\int 1/(25x^2+4) \mathrm{d} x$.
- practice Find $\int 1/(16x^2-9) \mathrm{d} x$.
- practice Find $\int 1/(x^2+2x+5) \mathrm{d} x$.
- practice Find $\int 1/(x^2-4x-5) \mathrm{d} x$.
- practice Find $\int 1/(x^2+6x+13) \mathrm{d} x$.
- practice Find $\int 1/(x^2-8x+7) \mathrm{d} x$.
- practice Find $\int 1/(x^2+4x+13) \mathrm{d} x$.
- practice Find $\int 1/(x^2-6x+5) \mathrm{d} x$.
- practice Find $\int 1/(x^2+10x+34) \mathrm{d} x$.
- practice Find $\int 1/(x^2-2x-8) \mathrm{d} x$.
- practice Find $\int 1/(x^2+8x+20) \mathrm{d} x$.
- practice Find $\int 1/(4x^2+4x+5) \mathrm{d} x$.
- practice Find $\int 1/(9x^2-12x-5) \mathrm{d} x$.
- practice Find $\int 1/(2x^2+4x+10) \mathrm{d} x$.
- practice Find $\int 1/(16x^2-8x-3) \mathrm{d} x$.
- practice $\int 1/(x^2+64) \mathrm{d} x$ equals
- practice $\int 1/(x^2-64) \mathrm{d} x$ equals
- practice Find $\int (x+2)/(2 x^2+6 x+5) \mathrm{d} x$.
- practice Find $\int (x+3)/\sqrt{x^2+4 x+5} \mathrm{d} x$.
- practice Find $\int 1/(8 - 2 x - x^2) \mathrm{d} x$.
- practice Find $\int (x + 1)/(x^2 + 4 x + 5) \mathrm{d} x$.
- practice Find $\int (x + 2)/\sqrt{x^2 + 2 x + 3} \mathrm{d} x$.
Answers
- $1/3 \cdot \tan^{-1}(x/3) + C$
- $1/4 \cdot \tan^{-1}(x/4) + C$
- $\tan^{-1}(x) + C$
- $1/6 \cdot \ln|(x-3)/(x+3)| + C$
- $1/10 \cdot \ln|(x-5)/(x+5)| + C$
- $1/2 \cdot \ln|(x-1)/(x+1)| + C$
- $1/6 \cdot \tan^{-1}(x/6) + C$
- $1/8 \cdot \ln|(x-4)/(x+4)| + C$
- $1/7 \cdot \tan^{-1}(x/7) + C$
- $1/6 \cdot \tan^{-1}(2x/3) + C$
- $1/12 \cdot \ln|(3x-2)/(3x+2)| + C$
- $1/10 \cdot \tan^{-1}(5x/2) + C$
- $1/24 \cdot \ln|(4x-3)/(4x+3)| + C$
- $1/2 \cdot \tan^{-1}((x+1)/2) + C$
- $1/6 \cdot \ln|(x-5)/(x+1)| + C$
- $1/2 \cdot \tan^{-1}((x+3)/2) + C$
- $1/6 \cdot \ln|(x-7)/(x-1)| + C$
- $1/3 \cdot \tan^{-1}((x+2)/3) + C$
- $1/4 \cdot \ln|(x-5)/(x-1)| + C$
- $1/3 \cdot \tan^{-1}((x+5)/3) + C$
- $1/6 \cdot \ln|(x-4)/(x+2)| + C$
- $1/2 \cdot \tan^{-1}((x+4)/2) + C$
- $1/4 \cdot \tan^{-1}(x+1/2) + C$
- $1/18 \cdot \ln|(3x-5)/(3x+1)| + C$
- $1/4 \cdot \tan^{-1}((x+1)/2) + C$
- $1/16 \cdot \ln|(4x-3)/(4x+1)| + C$
- (A) $1/8 \cdot \tan^{-1}(x/8) + C$.
- (A) $1/16 \cdot \ln|(x-8)/(x+8)| + C$.
- $(1/4) \ln|2 x^2+6 x+5| + (1/2) \tan^{-1}(2 x+3) + C$
- $\sqrt{x^2+4 x+5} + \ln|x+2+\sqrt{x^2+4 x+5}| + C$
- $(1/6) \ln|(x + 4)/(2 - x)| + C$
- $(1/2) \ln(x^2 + 4 x + 5) - \tan^{-1}(x + 2) + C$
- $\sqrt{x^2 + 2 x + 3} + \ln|x + 1 + \sqrt{x^2 + 2 x + 3}| + C$
Practice set: Exercise 7.4, part 2, and Exercise 7.7
These questions finish Exercise 7.4 and cover Exercise 7.7. They use the formulas with a square root, such as $1/\sqrt{a^2 - x^2}$ and $\sqrt{x^2 + a^2}$. Some need a substitution or a completed square first.
- practice Find $\int 1/\sqrt{9-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{16-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{25-x^2} \mathrm{d} x$.
- practice Find $\int 5/\sqrt{4-x^2} \mathrm{d} x$.
- practice Find $\int \sqrt{4-x^2} \mathrm{d} x$.
- practice Find $\int \sqrt{9-x^2} \mathrm{d} x$.
- practice Find $\int \sqrt{1-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2+9} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2+16} \mathrm{d} x$.
- practice Find $\int 3/\sqrt{x^2+4} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2+1} \mathrm{d} x$.
- practice Find $\int 4/\sqrt{1-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{4-9x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{9-4x^2} \mathrm{d} x$.
- practice Find $\int \sqrt{4-9x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2+2x+10} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{4x-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{2x-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2-6x+18} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{5-4x-x^2} \mathrm{d} x$.
- practice Find $\int \sqrt{8-2x-x^2} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{9-8x-x^2} \mathrm{d} x$.
- practice Which of these is $\int 1/\sqrt{4-x^2} \mathrm{d} x$?
- practice Which of these is $\int 1/\sqrt{x^2+4} \mathrm{d} x$?
- practice Find $\int \sqrt{x^2 + 4 x + 6} \mathrm{d} x$.
- practice Find $\int 1/\sqrt{x^2 - 9} \mathrm{d} x$.
- practice Find $\int \sqrt{x^2 - 4} \mathrm{d} x$.
Answers
- $\sin^{-1}(x/3) + C$
- $\sin^{-1}(x/4) + C$
- $\sin^{-1}(x/5) + C$
- $5 \cdot \sin^{-1}(x/2) + C$
- $x/2 \cdot \sqrt{4-x^2} + 2 \cdot \sin^{-1}(x/2) + C$
- $x/2 \cdot \sqrt{9-x^2} + 9/2 \cdot \sin^{-1}(x/3) + C$
- $x/2 \cdot \sqrt{1-x^2} + 1/2 \cdot \sin^{-1}(x) + C$
- $\ln|x+\sqrt{x^2+9}| + C$
- $\ln|x+\sqrt{x^2+16}| + C$
- $3 \cdot \ln|x+\sqrt{x^2+4}| + C$
- $\ln|x+\sqrt{x^2+1}| + C$
- $4 \cdot \sin^{-1}(x) + C$
- $1/3 \cdot \sin^{-1}(3x/2) + C$
- $1/2 \cdot \sin^{-1}(2x/3) + C$
- $x/2 \cdot \sqrt{4-9x^2} + 2/3 \cdot \sin^{-1}(3x/2) + C$
- $\ln|x+1+\sqrt{x^2+2x+10}| + C$
- $\sin^{-1}((x-2)/2) + C$
- $\sin^{-1}(x-1) + C$
- $\ln|x-3+\sqrt{x^2-6x+18}| + C$
- $\sin^{-1}((x+2)/3) + C$
- $(x+1)/2 \cdot \sqrt{8-2x-x^2} + 9/2 \cdot \sin^{-1}((x+1)/3) + C$
- $\sin^{-1}((x+4)/5) + C$
- (A) $\sin^{-1}(x/2) + C$.
- (A) $\ln|x+\sqrt{x^2+4}| + C$.
- $((x + 2)/2) \sqrt{x^2 + 4 x + 6} + \ln|x + 2 + \sqrt{x^2 + 4 x + 6}| + C$
- $\ln|x + \sqrt{x^2 - 9}| + C$
- $(x/2) \sqrt{x^2 - 4} - 2 \ln|x + \sqrt{x^2 - 4}| + C$
Practice set: Exercise 7.5
Exercise 7.5 practises partial fractions. Split the integrand, find the constants, then integrate each piece. A repeated factor or a quadratic factor needs a different split.
- practice Find $\int (5x+4)/((x-1)(x+2)) \mathrm{d} x$.
- practice Find $\int (5x-13)/((x-2)(x-5)) \mathrm{d} x$.
- practice Find $\int (x-11)/((x+3)(x-4)) \mathrm{d} x$.
- practice Find $\int 1/(x^2-9) \mathrm{d} x$.
- practice Find $\int (x+7)/((x+1)(x+4)) \mathrm{d} x$.
- practice Find $\int (4x-8)/((x-3)(x+1)) \mathrm{d} x$.
- practice Find $\int (x-1)/((x+1)(x+4)) \mathrm{d} x$.
- practice Find $\int (2x+3)/(x-1)^2 \mathrm{d} x$.
- practice Find $\int (5x-4)/(x+2)^2 \mathrm{d} x$.
- practice Find $\int (x-5)/(x+3)^2 \mathrm{d} x$.
- practice Find $\int 1/(x(x^2+9)) \mathrm{d} x$.
- practice Find $\int (x+8)/(x^2+x-6) \mathrm{d} x$.
- practice Find $\int 2/((x-1)(x-2)(x-3)) \mathrm{d} x$.
- practice Find $\int (x+1)/((x-1)(x-2)(x-3)) \mathrm{d} x$.
- practice Find $\int (5x+1)/((x-1)^2(x+2)) \mathrm{d} x$.
- practice Find $\int (10x-8)/((x-2)^2(x+1)) \mathrm{d} x$.
- practice Find $\int x/((x+3)(x^2+9)) \mathrm{d} x$.
- practice Find $\int 8/((x-2)(x^2+4)) \mathrm{d} x$.
- practice Which of these is $\int x/((x-1)(x-2)) \mathrm{d} x$?
- practice Which of these is $\int (3x-1)/(x+1)^2 \mathrm{d} x$?
- practice Find $\int (x^2+7x+4)/((x-1)(x+2)) \mathrm{d} x$.
- practice Find $\int (10-3x)/((x+2)(x^2+4)) \mathrm{d} x$.
- practice Find $\int (x^2+x+1)/((x-1)(x^2+1)) \mathrm{d} x$.
- practice Which of these is $\int 1/(x(x^2+4)) \mathrm{d} x$?
Answers
- $3 \cdot \ln|x-1| + 2 \cdot \ln|x+2| + C$
- $\ln|x-2| + 4 \cdot \ln|x-5| + C$
- $2 \cdot \ln|x+3| - \ln|x-4| + C$
- $1/6 \cdot \ln|x-3| - 1/6 \cdot \ln|x+3| + C$
- $2 \cdot \ln|x+1| - \ln|x+4| + C$
- $\ln|x-3| + 3 \cdot \ln|x+1| + C$
- $-2/3 \cdot \ln|x+1| + 5/3 \cdot \ln|x+4| + C$
- $2 \cdot \ln|x-1| - 5/(x-1) + C$
- $5 \cdot \ln|x+2| + 14/(x+2) + C$
- $\ln|x+3| + 8/(x+3) + C$
- $1/9 \cdot \ln|x| - 1/18 \cdot \ln(x^2+9) + C$
- $-\ln|x+3| + 2 \cdot \ln|x-2| + C$
- $\ln|x-1| - 2 \cdot \ln|x-2| + \ln|x-3| + C$
- $\ln|x-1| - 3 \cdot \ln|x-2| + 2 \cdot \ln|x-3| + C$
- $\ln|x-1| - 2/(x-1) - \ln|x+2| + C$
- $2 \cdot \ln|x-2| - 4/(x-2) - 2 \cdot \ln|x+1| + C$
- $-1/6 \cdot \ln|x+3| + 1/12 \cdot \ln(x^2+9) + 1/6 \cdot \tan^{-1}(x/3) + C$
- $\ln|x-2| - 1/2 \cdot \ln(x^2+4) - \tan^{-1}(x/2) + C$
- (B) $-\ln|x-1| + 2 \cdot \ln|x-2| + C$.
- (A) $3 \cdot \ln|x+1| + 4/(x+1) + C$.
- $x + 4 \cdot \ln|x-1| + 2 \cdot \ln|x+2| + C$
- $2 \cdot \ln|x+2| - \ln(x^2+4) + 1/2 \cdot \tan^{-1}(x/2) + C$
- $3/2 \cdot \ln|x-1| - 1/4 \cdot \ln(x^2+1) + 1/2 \cdot \tan^{-1}(x) + C$
- (A) $1/4 \cdot \ln|x| - 1/8 \cdot \ln(x^2+4) + C$.
Practice set: Exercise 7.6
Exercise 7.6 practises integration by parts. Some questions need it twice. Others fit the formula $\int e^x (f(x) + f'(x)) \mathrm{d} x = e^x f(x) + C$, so look for that first.
- practice Find $\int x \cdot \sin 2 x \mathrm{d} x$.
- practice Find $\int x \cdot e^{3 x} \mathrm{d} x$.
- practice Find $\int x \cdot \cos x \mathrm{d} x$.
- practice Find $\int \ln x \mathrm{d} x$.
- practice Find $\int (x+1) \cdot e^x \mathrm{d} x$.
- practice Find $\int \tan^{-1}(2 x) \mathrm{d} x$.
- practice Find $\int x \cdot \ln(3 x) \mathrm{d} x$.
- practice Find $\int (2 x+1) \cdot \ln x \mathrm{d} x$.
- practice Find $\int x \cdot e^{-2 x} \mathrm{d} x$.
- practice Which of these equals $\int x \cdot e^{-x} \mathrm{d} x$?
- practice Find $\int x^2 \cdot e^{2 x} \mathrm{d} x$.
- practice Find $\int x^2 \cdot \sin x \mathrm{d} x$.
- practice Find $\int x^3 \cdot \ln x \mathrm{d} x$.
- practice Find $\int (\ln x)^2 \mathrm{d} x$.
- practice Find $\int x \cdot \cos^{-1}(x) \mathrm{d} x$.
- practice Find $\int e^x (\sqrt{x} + 1/(2 \sqrt{x})) \mathrm{d} x$.
- practice Find $\int e^x (\cos x - \sin x) \mathrm{d} x$.
- practice Find $\int e^x (x^3+3 x^2) \mathrm{d} x$.
- practice Find $\int e^{3 x} \cdot \cos x \mathrm{d} x$.
- practice Which of these equals $\int e^x (\ln x + 1/x) \mathrm{d} x$?
- practice Find $\int x^2 e^{-x} \mathrm{d} x$.
- practice Find $\int x \sec^2 x \mathrm{d} x$.
- practice Find $\int x \cos(3 x) \mathrm{d} x$.
- practice Find $\int \sin^{-1}(x) \mathrm{d} x$.
Answers
- $-(x \cos 2 x)/2 + (\sin 2 x)/4 + C$
- $(x e^{3 x})/3 - e^{3 x}/9 + C$
- $x \sin x + \cos x + C$
- $x \ln x - x + C$
- $x e^x + C$
- $x \tan^{-1}(2 x) - (1/4) \ln(1+4 x^2) + C$
- $(x^2/2) \ln(3 x) - x^2/4 + C$
- $(x^2+x) \ln x - x^2/2 - x + C$
- $-(x e^{-2 x})/2 - e^{-2 x}/4 + C$
- (A) $-x e^{-x} - e^{-x} + C$.
- $(x^2 e^{2 x})/2 - (x e^{2 x})/2 + e^{2 x}/4 + C$
- $-x^2 \cos x + 2 x \sin x + 2 \cos x + C$
- $(x^4/4) \ln x - x^4/16 + C$
- $x (\ln x)^2 - 2 x \ln x + 2 x + C$
- $(x^2 \cos^{-1}(x))/2 + (\sin^{-1}(x))/4 - (x \sqrt{1-x^2})/4 + C$
- $e^x \sqrt{x} + C$
- $e^x \cos x + C$
- $e^x x^3 + C$
- $(e^{3 x}(\sin x+3 \cos x))/10 + C$
- (A) $e^x \ln x + C$.
- $-e^{-x} (x^2+2 x+2) + C$
- $x \tan x + \ln|\cos x| + C$
- $(x \sin(3 x))/3 + (\cos(3 x))/9 + C$
- $x \sin^{-1}(x) + \sqrt{1-x^2} + C$
Practice set: the definite integral from strips (beyond the board syllabus)
These questions find a definite integral straight from the strips, as a limit of totals. The board exam does not set these questions, and NCERT no longer has this exercise. Try a few to see the work the fundamental theorem saves.
- practice Evaluate $\int_1^4 5 \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^2 x \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_1^3 x \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^1 2 x \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^2 (3 x+1) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_1^3 (x+2) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_{-1}^1 x \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^4 (5-x) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^1 x^2 \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^3 x^2 \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_1^2 x^2 \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Which value does $\int_{-1}^2 x^2 \mathrm{d} x$ tend to, once its limit is taken?
- practice Evaluate $\int_0^2 (x^2+1) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^2 (2 x^2-x) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^3 (x^2+2 x) \mathrm{d} x$ directly from the definition, as a limit of totals.
- practice Evaluate $\int_0^1 x^3 \mathrm{d} x$ directly from the definition, as a limit of totals.
Answers
- $15$
- $2$
- $4$
- $1$
- $8$
- $8$
- $0$
- $12$
- $1/3$
- $9$
- $7/3$
- (A) $3$.
- $14/3$
- $10/3$
- $18$
- $1/4$
Practice set: Exercise 7.8
Exercise 7.8 practises the second fundamental theorem. Find one integral, then subtract its value at the lower limit from its value at the upper limit. Some need a method from earlier, such as substitution or by parts.
- practice Evaluate $\int_0^2 (x+3) \mathrm{d} x$.
- practice Evaluate $\int_1^3 1/x \mathrm{d} x$.
- practice Evaluate $\int_1^2 (4 x^3-3 x^2+2 x+1) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sin x \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/4} \cos 2 x \mathrm{d} x$.
- practice Evaluate $\int_2^3 e^x \mathrm{d} x$.
- practice Evaluate $\int_0^1 x/(1+x^2) \mathrm{d} x$.
- practice Evaluate $\int_3^4 1/((x-1)(x-2)) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x \cdot e^x \mathrm{d} x$.
- practice Which value does $\int_0^1 1/(1+x^2) \mathrm{d} x$ equal?
- practice Evaluate $\int_0^{1/2} 1/\sqrt{1-x^2} \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/4} (\sec^2 x+x) \mathrm{d} x$.
- practice Evaluate $\int_0^1 (3 x^2+2 x) \mathrm{d} x$.
- practice Evaluate $\int_2^4 1/x^2 \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/3} \sec^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^1 x^2 e^x \mathrm{d} x$.
- practice Evaluate $\int_0^2 1/(x^2+4) \mathrm{d} x$.
- practice Evaluate $\int_0^1 1/(x^2-4) \mathrm{d} x$.
- practice Evaluate $\int_0^{\sqrt{3}} 1/\sqrt{4-x^2} \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} x \cos x \mathrm{d} x$.
- practice Evaluate $\int_1^9 1/\sqrt{x} \mathrm{d} x$.
- practice Evaluate $\int_2^3 1/(x^2-1) \mathrm{d} x$.
Answers
- $8$
- $\ln 3$
- $12$
- $1$
- $1/2$
- $e^3-e^2$
- $(\ln 2)/2$
- $\ln(4/3)$
- $1$
- (A) $\pi/4$.
- $\pi/6$
- $1+\pi^2/32$
- $2$
- $1/4$
- $\sqrt{3}$
- $e-2$
- $\pi/8$
- $-(1/4) \ln 3$
- $\pi/3$
- $\pi/2-1$
- $4$
- $(1/2) \ln(3/2)$
Practice set: Exercise 7.9
Exercise 7.9 practises substitution in definite integrals. Change the limits in the same step as the variable. Then stay in the new variable to the end.
- practice Evaluate $\int_0^1 (2 x)/(x^2+1) \mathrm{d} x$.
- practice Evaluate $\int_0^2 (3 x^2)/(x^3+1) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sin x \cos^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/4} \sec^2 x \tan x \mathrm{d} x$.
- practice Evaluate $\int_0^1 x e^{x^2} \mathrm{d} x$.
- practice Evaluate $\int_1^2 (2 x+1)/(x^2+x+1) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x/(x^2+1)^2 \mathrm{d} x$.
- practice Evaluate $\int_1^e (\ln x)/x \mathrm{d} x$.
- practice Evaluate $\int_0^1 x^2/(x^3+8) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} (\cos x)/(1+\sin^2 x) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x \sqrt{1-x^2} \mathrm{d} x$.
- practice Evaluate $\int_0^1 x^3/(x^4+1) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x(1-x)^4 \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/4} \tan^3 x \sec^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^1 (x+1)/\sqrt{x^2+2 x+3} \mathrm{d} x$.
- practice Evaluate $\int_e^{e^2} 1/(x \ln x) \mathrm{d} x$.
- practice The value of $\int_0^{\pi/2} (\sin x)/(1+\cos^2 x) \mathrm{d} x$ is
- practice The value of $\int_0^1 (2 x+3)/(x^2+3 x+1) \mathrm{d} x$ is
- practice Evaluate $\int_0^1 (3 x^2-2 x)/(x^3-x^2+1) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x^3/\sqrt{1+x^4} \mathrm{d} x$.
Answers
- $\ln 2$
- $\ln 9$
- $1/3$
- $1/2$
- $(1/2) (e-1)$
- $\ln(7/3)$
- $1/4$
- $1/2$
- $(1/3) \ln(9/8)$
- $\pi/4$
- $1/3$
- $(1/4) \ln 2$
- $1/30$
- $1/4$
- $\sqrt{6} - \sqrt{3}$
- $\ln 2$
- (A) $\pi/4$.
- (A) $\ln 5$.
- $0$
- $(1/2) (\sqrt{2}-1)$
Practice set: Exercise 7.10 and Miscellaneous
Exercise 7.10 practises the properties of definite integrals. The Miscellaneous Exercise then mixes every method you have met. For an area, split the interval where the graph crosses the axis.
- practice If $\int_2^5 f(x) \mathrm{d} x = 7$, find $\int_5^2 f(x) \mathrm{d} x$.
- practice If $\int_1^4 g(x) \mathrm{d} x = 10$, evaluate $\int_4^1 3 g(x) \mathrm{d} x$.
- practice Evaluate $\int_{-\pi/2}^{\pi/2} \sin^3 x \mathrm{d} x$.
- practice Evaluate $\int_{-1}^1 x^3 \cos x \mathrm{d} x$.
- practice Evaluate $\int_{-3}^3 (x^5+2 x^3-x) \mathrm{d} x$.
- practice Evaluate $\int_{-2}^2 x^4 \mathrm{d} x$.
- practice Given that $\sin^2 x$ has period $\pi$, evaluate $\int_0^{6 \pi} \sin^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sin^2 x \mathrm{d} x$ using the reflection $x \to \pi/2 - x$.
- practice Evaluate $\int_0^1 \ln((1-x)/x) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sqrt{\sin x}/(\sqrt{\sin x}+\sqrt{\cos x}) \mathrm{d} x$.
- practice Evaluate $\int_0^\pi x (\sin x)/(1+\cos^2 x) \mathrm{d} x$.
- practice Evaluate $\int_0^\pi x \sin^3 x \mathrm{d} x$.
- practice Show that $\int_0^1 x(1-x)^n \mathrm{d} x = \int_0^1 x^n \cdot (1-x) \mathrm{d} x$ for any positive integer $n$.
- practice Show that $\int_{-a}^a f(x) g(x) \mathrm{d} x = 0$ if $f$ is even and $g$ is odd.
- practice The value of $\int_{-2}^2 (x^5+x \cos 3 x+\sin^3 x) \mathrm{d} x$ is
- practice If $\int_0^a 1/(4+x^2) \mathrm{d} x = \pi/8$, then $a$ equals
- practice Evaluate the definite integral $\int_0^3 (x-1) \mathrm{d} x$.
- practice The line $y=x-1$ crosses the $x$-axis at $x=1$. Find the true area enclosed between the line and the $x$-axis on $[0,3]$.
- practice True or false, with a reason: a definite integral that evaluates to $0$ means the function is zero everywhere on the interval.
- practice Evaluate $\int_0^1 x e^x \mathrm{d} x$.
- practice Evaluate $\int_2^3 (3 x+3)/((x-1)(x+2)) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sin^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^2 1/(x^2+4) \mathrm{d} x$.
- practice Evaluate $\int_1^2 (x^2+1)/x \mathrm{d} x$.
- practice The value of $\int_0^1 (2 x+1) e^{x^2+x} \mathrm{d} x$ is
- practice The value of $\int_{-1}^1 \left|x\right| \mathrm{d} x$ is
- practice Prove that $\int_{-\pi}^\pi x^3 \sin^2 x \mathrm{d} x = 0$.
- practice Prove that $\int_0^{\pi/2} \sin^3 x \mathrm{d} x = 2/3$.
- practice Prove that $\int_0^1 1/((x+1)(x+2)) \mathrm{d} x = \ln(4/3)$.
- practice Evaluate $\int_0^{\pi/4} \tan x \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} 1/(1+\tan x) \mathrm{d} x$.
- practice Evaluate $\int_0^1 e^x (1+x) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/4} \sin^2 x \cos^2 x \mathrm{d} x$.
- practice Evaluate $\int_0^1 \sqrt{1-x^2} \mathrm{d} x$.
- practice Evaluate $\int_0^1 1/(x^2-4) \mathrm{d} x$.
- practice Evaluate $\int_0^1 (2 x)/((x^2+1)(x^2+2)) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} \sin 2 x/(1+\sin^2 x) \mathrm{d} x$.
- practice Evaluate $\int_0^1 x^2 e^x \mathrm{d} x$.
- practice Evaluate $\int_0^1 1/((x+1)(x^2+1)) \mathrm{d} x$.
- practice Evaluate $\int_0^{\pi/2} x \sin x \cos x \mathrm{d} x$.
- practice Evaluate $\int_0^{2 \pi} \cos^5 x \mathrm{d} x$.
- practice Evaluate $\int_0^4 \left|x - 1\right| \mathrm{d} x$.
Answers
- $-7$
- $-30$
- $0$
- $0$
- $0$
- $64/5$
- $3 \pi$
- $\pi/4$
- $0$
- $\pi/4$
- $\pi^2/4$
- $2 \pi/3$
- Shown: by the reflection property $\int_a^b f(x) \mathrm{d} x = \int_a^b f(a+b-x) \mathrm{d} x$ with $a=0$, $b=1$; $a+b-x=1-x$ turns $x(1-x)^n$ into $(1-x) x^n$.
- Shown: an even function times an odd function is odd, and the integral of an odd function over an interval symmetric about $0$ is $0$.
- (A) $0$.
- (B) $2$.
- $3/2$
- $5/2$
- False: positive and negative contributions can cancel to give a net integral of $0$ even though $f$ is not the zero function on the interval.
- $1$
- $\ln 5$
- $\pi/4$
- $\pi/8$
- $3/2 + \ln 2$
- (A) $e^2-1$.
- (B) $1$.
- True: $x^3$ is odd and $\sin^2 x$ is even, so the product is odd, and an odd function integrates to $0$ over an interval symmetric about $0$.
- True: using $\sin^3 x = \sin x(1-\cos^2 x)$ and the substitution $t=\cos x$.
- True: the partial-fraction split $1/((x+1)(x+2)) = 1/(x+1) - 1/(x+2)$, integrated and evaluated at the limits, gives $\ln(4/3)$.
- $(1/2) \ln 2$
- $\pi/4$
- $e$
- $\pi/32$
- $\pi/4$
- $-(1/4) \ln 3$
- $\ln(4/3)$
- $\ln 2$
- $e-2$
- $(1/4) \ln 2 + \pi/8$
- $\pi/8$
- $0$
- $5$
- practice $\int e^x (\tan x + \ln(\sec x)) \mathrm{d} x$ equals ______. (1 mark)
- practice Assertion (A). $\int_{-\pi/2}^{\pi/2} \sin x \mathrm{d} x = 0$. Reason (R). $\sin x$ is an odd function, and $\int_{-a}^a f(x) \mathrm{d} x = 0$ whenever $f$ is odd. (1 mark)
- practice Find $\int \sec^2 x/(1+\tan x) \mathrm{d} x$. (2 marks)
- practice Find $a > 0$ such that $\int_0^a (2 x+1) \mathrm{d} x = 12$. (3 marks)
- practice Evaluate $\int 1/((x-1)(x^2+1)) \mathrm{d} x$. (5 marks)
- practice A firm’s marginal cost of producing $x$ units is $C'(x) = 3 x^2-24 x+50$ rupees per unit, and the fixed cost is $C(0) = 500$ rupees. (i) Find $C(x)$. (1 mark) (ii) Find the cost of producing $10$ units. (1 mark) (iii) Find the extra cost of producing units $11$ to $20$. (2 marks)
- practice Evaluate $\int_0^{\pi/2} \sin x/(\sin x+\cos x) \mathrm{d} x$. (3 marks)
Answers
- (B) $e^x \ln(\sec x) + C$.
- (A) Both A and R are true and R is the correct explanation of A.
- $\ln|1+\tan x| + C$.
- $a = 3$.
- $(1/2) \ln|x-1| - (1/4) \ln(x^2+1) - (1/2) \tan^{-1}(x) + C$.
- (i) $C(x) = x^3-12 x^2+50 x+500$. (ii) ₹$800$. (iii) ₹$3900$.
- $\pi/4$.