Antiderivatives

A car is travelling at $20$ m/s when the driver brakes. From that moment, its velocity after $t$ seconds is $v(t)=20-5t$ (m/s), and the car comes to a stop after $4$ seconds. How many metres does it travel in those $4$ seconds?

You might think: $20\cdot 4=80$ m. But the car keeps slowing down, so the distance must be less. To answer, we need to run differentiation backwards.

What is an antiderivative?

Explore: which functions have derivative $2x$ ?

In grade 11 we learnt to differentiate. Now we ask the reverse question: if the derivative is $2x$ , what was the function?

Function $F(x)$	$x^2$	$x^2+1$	$x^2-5$	$x^2+\sqrt3$
Derivative $F'(x)$	$2x$	$2x$	$2x$	$2x$

There are many answers, but they differ only by a constant, because the derivative of a constant is $0$ . In the car problem, the velocity is the derivative of the distance, so we need a function $s(t)$ with $s'(t)=20-5t$ .

Why is it true? If $G(x)$ is another antiderivative, then $\big(G(x)-F(x)\big)'=f(x)-f(x)=0$ on $K$ . A function whose derivative is $0$ on a whole interval neither increases nor decreases, so it is a constant: $G(x)-F(x)=C$ .

The symbol $\displaystyle\int f(x)\,dx$ is read "the indefinite integral of $f$ "; it stands for the whole family of antiderivatives. Every function that is continuous on $K$ has an antiderivative on $K$ . The definition also gives $\displaystyle\int f'(x)\,dx=f(x)+C$ .

The table and the rules

Each row of this table is a derivative formula read backwards. You can check every row by differentiating the right-hand side.

Function $f(x)$	Indefinite integral $\displaystyle\int f(x)\,dx$
$0$	$C$
$1$	$x+C$
$x^{\alpha}$ $(\alpha\neq-1)$	$\dfrac{x^{\alpha+1}}{\alpha+1}+C$
$\dfrac{1}{x}$	$\ln\lvert x\rvert+C$
$e^x$	$e^x+C$
$a^x$ $(0<a\neq1)$	$\dfrac{a^x}{\ln a}+C$
$\cos x$	$\sin x+C$
$\sin x$	$-\cos x+C$
$\dfrac{1}{\cos^2x}$	$\tan x+C$
$\dfrac{1}{\sin^2x}$	$-\cot x+C$

Notice the minus sign in the $\sin x$ row: since $(\cos x)'=-\sin x$ , to get $\sin x$ we must take $-\cos x$ .

Both rules hold because differentiation follows the same rules. With them, we find the antiderivative of a sum one term at a time.

With roots and fractions, we rewrite them as powers before using the table.

Example Roots and fractions

Find $\displaystyle\int\left(2\sqrt x-\frac3x+\frac1{x^2}\right)dx$ on $(0, +\infty)$ .

Solution. We write $\sqrt x=x^{\frac12}$ and $\dfrac{1}{x^2}=x^{-2}$ . The term $\dfrac3x$ has exponent $-1$ , so it uses the $\ln\lvert x\rvert$ formula:

$\begin{aligned} \int\left(2x^{\frac12}-\frac3x+x^{-2}\right)dx&=2\cdot\frac{x^{\frac32}}{\frac32}-3\ln\lvert x\rvert+\frac{x^{-1}}{-1}+C\\ &=\frac43x\sqrt x-3\ln x-\frac1x+C. \end{aligned}$

On $(0, +\infty)$ we have $\lvert x\rvert=x$ , so we may write $\ln x$ .

Next come exponential and trigonometric functions. Watch every sign.

Many functions are not in the table, but a small rearrangement puts them there.

Example Rearranging first

Find a) $\displaystyle\int\frac{(x+1)^2}{x}\,dx$ on $(0, +\infty)$ ; b) $\displaystyle\int\tan^2x\,dx$ on $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ .

Solution. a) We expand the numerator and divide each term by $x$ :

$\int\frac{x^2+2x+1}{x}\,dx=\int\left(x+2+\frac1x\right)dx=\frac{x^2}{2}+2x+\ln x+C.$

b) From $1+\tan^2x=\dfrac{1}{\cos^2x}$ we get $\tan^2x=\dfrac{1}{\cos^2x}-1$ :

$\int\tan^2x\,dx=\int\left(\frac{1}{\cos^2x}-1\right)dx=\tan x-x+C.$

The antiderivative that meets a condition

A family of antiderivatives has infinitely many members. When we are given one more value, such as $F(0)=5$ , we substitute it to find exactly one constant $C$ .

Graphically, each constant $C$ gives one graph, and these graphs are vertical translations of one another. The condition $F(x_0)=y_0$ picks out the one graph through the point $(x_0, y_0)$ .

Three antiderivatives of

f(x)=2x

From velocity to distance

If $s(t)$ is the distance an object has travelled after $t$ seconds, its velocity is $v(t)=s'(t)$ . So $s(t)$ is an antiderivative of $v(t)$ . In the same way, the velocity is an antiderivative of the acceleration. Now we can answer the question about the car.

Watch out Common mistakes

Using the power rule for $\dfrac1x$ . Wrong: $\displaystyle\int x^{-1}dx=\frac{x^0}{0}+C$ . Why it is wrong: the formula $\dfrac{x^{\alpha+1}}{\alpha+1}$ only works when $\alpha\neq-1$ . Right: $\displaystyle\int\frac1x\,dx=\ln\lvert x\rvert+C$ .
The wrong sign with $\sin x$ . Wrong: $\displaystyle\int\sin x\,dx=\cos x+C$ . Why it is wrong: $(\cos x)'=-\sin x$ . Right: $\displaystyle\int\sin x\,dx=-\cos x+C$ .
Splitting the antiderivative of a product. Wrong: $\displaystyle\int x\cdot x\,dx=\frac{x^2}{2}\cdot\frac{x^2}{2}+C$ . Why it is wrong: there is no rule "the antiderivative of a product is the product of the antiderivatives"; differentiating the right-hand side gives $x^3\neq x^2$ . Right: $\displaystyle\int x^2\,dx=\frac{x^3}{3}+C$ .
Taking $e^0=0$ when finding $C$ . Wrong: $F(x)=e^x+2x^2+C$ and $F(0)=5$ , so $C=5$ . Why it is wrong: $e^0=1$ . Right: $1+C=5$ , so $C=4$ .

Key terms

English	Tiếng Việt
antiderivative	nguyên hàm
family of antiderivatives	họ nguyên hàm
indefinite integral	tích phân bất định
arbitrary constant $C$	hằng số tuỳ ý $C$
table of standard antiderivatives	bảng nguyên hàm cơ bản
velocity	vận tốc
acceleration	gia tốc
distance travelled	quãng đường

Summary

$F$ is an antiderivative of $f$ on $K$ when $F'(x)=f(x)$ for every $x\in K$ .
Every antiderivative has the form $F(x)+C$ ; we write $\displaystyle\int f(x)\,dx=F(x)+C$ .
The table to remember: $x^\alpha\to\dfrac{x^{\alpha+1}}{\alpha+1}$ ( $\alpha\neq-1$ ), $\dfrac1x\to\ln\lvert x\rvert$ , $e^x\to e^x$ , $a^x\to\dfrac{a^x}{\ln a}$ , $\cos x\to\sin x$ , $\sin x\to-\cos x$ , $\dfrac1{\cos^2x}\to\tan x$ , $\dfrac1{\sin^2x}\to-\cot x$ .
Constant factors, sums and differences split apart; products do not.
Given $F(x_0)=y_0$ , substitute it to find $C$ .
Distance is an antiderivative of velocity; always check by differentiating.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Find $\displaystyle\int(4x^3-6x+7)\,dx$ .

What you need and how to start

What you need: $\displaystyle\int x^\alpha dx=\frac{x^{\alpha+1}}{\alpha+1}+C$ and the rule for sums.

Steps:

Find the antiderivative of each term.
Add a constant $C$ and check by differentiating.

Show solution

$\displaystyle\int(4x^3-6x+7)\,dx=4\cdot\frac{x^4}{4}-6\cdot\frac{x^2}{2}+7x+C=x^4-3x^2+7x+C$ .

Answer. $x^4-3x^2+7x+C$ .

Exercise 2

Find $\displaystyle\int\left(3\sqrt x+\frac1{x^2}\right)dx$ on $(0, +\infty)$ .

What you need and how to start

What you need: write $\sqrt x=x^{\frac12}$ and $\dfrac{1}{x^2}=x^{-2}$ , then use the power formula.

Steps:

Rewrite as powers.
Apply the formula to each term.

Show solution

$\displaystyle\int\left(3x^{\frac12}+x^{-2}\right)dx=3\cdot\frac{x^{\frac32}}{\frac32}+\frac{x^{-1}}{-1}+C=2x\sqrt x-\frac1x+C$ .

Answer. $2x\sqrt x-\dfrac1x+C$ .

Exercise 3

Find $\displaystyle\int\left(5e^x-2\sin x+\frac{1}{\sin^2x}\right)dx$ .

What you need and how to start

What you need: the $e^x$ , $\sin x$ and $\dfrac{1}{\sin^2x}$ rows of the table.

Steps:

Write the antiderivative of each term, watching the two minus signs.
Put them together and add $C$ .

Show solution

$\displaystyle\int\left(5e^x-2\sin x+\frac{1}{\sin^2x}\right)dx=5e^x-2(-\cos x)+(-\cot x)+C=5e^x+2\cos x-\cot x+C$ .

Answer. $5e^x+2\cos x-\cot x+C$ .

Exercise 4

Find $\displaystyle\int(2x-1)^2\,dx$ .

What you need and how to start

What you need: do not find the antiderivative of a square by "squaring the antiderivative"; expand first.

Steps:

Expand $(2x-1)^2=4x^2-4x+1$ .
Find the antiderivative of the polynomial.

Show solution

$\displaystyle\int(4x^2-4x+1)\,dx=\frac{4x^3}{3}-2x^2+x+C$ .

Answer. $\dfrac43x^3-2x^2+x+C$ .

Exercise 5

Consider $f(x)=3x^2+2\cos x$ . Is each claim true or false?

a) $F(x)=x^3+2\sin x$ is an antiderivative of $f(x)$ .

b) $F(x)=x^3-2\sin x+5$ is an antiderivative of $f(x)$ .

c) The antiderivative $G(x)$ of $f(x)$ with $G(0)=1$ is $G(x)=x^3+2\sin x+1$ .

d) For $G(x)$ in c), $G\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^3}{8}+2$ .

What you need and how to start

What you need: $F$ is an antiderivative of $f$ when $F'=f$ ; the condition $G(0)=1$ fixes the constant $C$ .

Steps:

Differentiate the functions in a) and b).
Find $C$ in c), then substitute $x=\dfrac{\pi}{2}$ for d).

Show solution

a) True: $(x^3+2\sin x)'=3x^2+2\cos x$ .

b) False: $(x^3-2\sin x+5)'=3x^2-2\cos x\neq f(x)$ .

c) True: $G(x)=x^3+2\sin x+C$ and $G(0)=C=1$ .

d) False: $G\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^3}{8}+2\cdot1+1=\dfrac{\pi^3}{8}+3$ .

Answer. a) True; b) False; c) True; d) False.

Exercise 6

$F(x)$ is the antiderivative of $f(x)=4x^3-2x$ with $F(1)=3$ . Find $F(2)$ .

What you need and how to start

What you need: find the family of antiderivatives, then use $F(1)=3$ to find $C$ .

Steps:

$F(x)=x^4-x^2+C$ .
Substitute $x=1$ to find $C$ , then compute $F(2)$ .

Show solution

$F(1)=1-1+C=3$ , so $C=3$ . Then $F(2)=16-4+3=15$ .

Answer. $15$ .

Exercise 7

A tank already holds $200$ litres of water. A pump is switched on, and $t$ minutes later the volume of water is changing at the rate $V'(t)=6t+20$ (litres per minute). How many litres are in the tank after $10$ minutes?

What you need and how to start

What you need: the volume $V(t)$ is an antiderivative of $V'(t)$ ; at the start, $V(0)=200$ .

Steps:

Find $V(t)=3t^2+20t+C$ .
Use $V(0)=200$ to find $C$ , then compute $V(10)$ .

Show solution

$V(0)=C=200$ , so $V(t)=3t^2+20t+200$ and $V(10)=300+200+200=700$ .

Answer. $700$ litres.

Exercise 8 Review

An object moves in a straight line with velocity $v(t)=12t-3t^2$ (m/s), $0\le t\le 4$ . Find its greatest velocity and the distance it travels in the first $4$ seconds.

What you need and how to start

What you need: how to find a maximum on a closed interval (see maximum and minimum values); distance is an antiderivative of velocity.

Steps:

Solve $v'(t)=0$ and compare $v(0)$ , $v(2)$ , $v(4)$ .
Find $s(t)$ with $s(0)=0$ , then compute $s(4)$ .

Show solution

$v'(t)=12-6t=0\iff t=2$ ; $v(0)=0$ , $v(2)=12$ , $v(4)=0$ . Since $v(t)\ge0$ on $[0, 4]$ , the object never turns back, and $s(t)=6t^2-t^3$ (because $s(0)=0$ ), so $s(4)=96-64=32$ .

Answer. Greatest velocity $12$ m/s (at $t=2$ ); distance $32$ m.

Exercise 9 Challenge

Find the function $f(x)$ with $f''(x)=6x+2$ , $f'(0)=1$ and $f(0)=-1$ .

What you need and how to start

What you need: $f'$ is an antiderivative of $f''$ , and $f$ is an antiderivative of $f'$ (second derivatives were met in grade 11).

Steps:

Find $f'(x)$ and use $f'(0)=1$ to find the first constant.
Find $f(x)$ and use $f(0)=-1$ to find the second constant.

Show solution

$f'(x)=3x^2+2x+C_1$ and $f'(0)=C_1=1$ . Next, $f(x)=x^3+x^2+x+C_2$ and $f(0)=C_2=-1$ . Check: $f''(x)=6x+2$ .

Answer. $f(x)=x^3+x^2+x-1$ .

Exercise 10 Challenge

Find $\displaystyle\int\sin^2\frac{x}{2}\,dx$ .

What you need and how to start

What you need: the double-angle formula (grade 11): $\cos x=1-2\sin^2\dfrac x2$ , so $\sin^2\dfrac x2=\dfrac{1-\cos x}{2}$ .

Steps:

Replace $\sin^2\dfrac x2$ by $\dfrac12-\dfrac12\cos x$ .
Find the antiderivative term by term.

Show solution

$\displaystyle\int\left(\frac12-\frac12\cos x\right)dx=\frac x2-\frac12\sin x+C$ . Check: $\left(\dfrac x2-\dfrac12\sin x\right)'=\dfrac{1-\cos x}{2}=\sin^2\dfrac x2$ .

Answer. $\dfrac x2-\dfrac{\sin x}{2}+C$ .

What is an antiderivative?

Explore: which functions have derivative $2x$ ?

The table and the rules

The antiderivative that meets a condition

From velocity to distance

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $F(x)=e^x-x^2$ is an antiderivative of $f(x)$ .	True	False
b) $\displaystyle\int f(x)\,dx=e^x-2+C$ .	True	False
c) The antiderivative $F(x)$ of $f(x)$ with $F(0)=3$ is $F(x)=e^x-x^2+2$ .	True	False
d) For $F(x)$ in c), $F(1)=e+1$ .	True	False

Antiderivatives

What is an antiderivative?

Explore: which functions have derivative 2x2x2x?

The table and the rules

The antiderivative that meets a condition

From velocity to distance

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Explore: which functions have derivative $2x$ ?