Revision: antiderivatives and integrals

Antiderivatives and integrals usually take two or three questions in Part I (standard results, properties), possibly a true/false question in Part II and an area or real-life question in Part III, altogether usually about $1$ to $2$ marks. A scientific calculator can evaluate definite integrals, so this is the easiest part of the paper to check.

The details are in the lessons on antiderivatives, definite integrals and applications of integrals.

Antiderivatives

Note Standard antiderivatives (

C

is a constant)

$f(x)$	$\displaystyle\int f(x)\,dx$
$x^\alpha$ ( $\alpha\ne-1$ )	$\dfrac{x^{\alpha+1}}{\alpha+1}+C$
$\dfrac1x$	$\ln\lvert x\rvert+C$
$e^x$	$e^x+C$
$a^x$ ( $0<a\ne1$ )	$\dfrac{a^x}{\ln a}+C$
$\cos x$	$\sin x+C$
$\sin x$	$-\cos x+C$
$\dfrac{1}{\cos^2x}$	$\tan x+C$
$\dfrac{1}{\sin^2x}$	$-\cot x+C$

Properties: $\displaystyle\int\left[f(x)\pm g(x)\right]dx=\int f(x)\,dx\pm\int g(x)\,dx$ and $\displaystyle\int kf(x)\,dx=k\int f(x)\,dx$ ( $k\ne0$ ).

The quickest check: differentiate your result; you must get $f(x)$ back.

Example Multiple choice: an antiderivative with a condition

$F(x)$ is an antiderivative of $f(x)=3x^2+2\sin x$ with $F(0)=1$ . Then $F(x)$ is A. $x^3-2\cos x+3$ . B. $x^3+2\cos x-1$ . C. $x^3-2\cos x+1$ . D. $6x+2\cos x+3$ .

Solution. We use the standard results, noting that $\displaystyle\int\sin x\,dx=-\cos x+C$ :

$F(x)=x^3-2\cos x+C,\qquad F(0)=0-2\cos0+C=-2+C=1\iff C=3.$

So $F(x)=x^3-2\cos x+3$ : the answer is A. Option B has the wrong sign for the antiderivative of $\sin x$ ; C forgets that $\cos0=1$ ; D differentiates instead of integrating.

Definite integrals

Theorem The Newton–Leibniz formula and properties

If $F$ is an antiderivative of a function $f$ continuous on $[a, b]$ , then $\displaystyle\int_a^b f(x)\,dx=\Big[F(x)\Big]_a^b=F(b)-F(a)$ . For continuous $f$ , $g$ and $c\in[a, b]$ :

$\displaystyle\int_a^b\left[f(x)\pm g(x)\right]dx=\int_a^b f(x)\,dx\pm\int_a^b g(x)\,dx$ , $\displaystyle\int_a^b kf(x)\,dx=k\int_a^b f(x)\,dx$ ;
$\displaystyle\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx$ ;
$\displaystyle\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$ .

Example True/false: using the properties of integrals

The function $f$ is continuous on $\mathbb{R}$ , with $\displaystyle\int_0^3 f(x)\,dx=5$ and $\displaystyle\int_0^1 f(x)\,dx=2$ . a) $\displaystyle\int_1^3 f(x)\,dx=3$ . b) $\displaystyle\int_0^3\left[2f(x)-1\right]dx=7$ . c) $\displaystyle\int_3^1 f(x)\,dx=3$ . d) $\displaystyle\int_0^1\left[f(x)+x\right]dx=2.5$ .

Solution. a) True: split $[0, 3]$ at $1$ : $\displaystyle\int_1^3 f=5-2=3$ .

b) True: $2\cdot5-\displaystyle\int_0^3 1\,dx=10-3=7$ .

c) False: swapping the limits changes the sign, so $\displaystyle\int_3^1 f=-3$ .

d) True: $2+\displaystyle\int_0^1 x\,dx=2+\dfrac12=2.5$ .

Key: T T F T.

Area and volume

To remove the modulus, find the roots of $f(x)-g(x)=0$ in $[a, b]$ , split the interval at those roots, and on each piece take "upper curve minus lower curve".

The region between

y=2x

and

y=x^2

Real-life applications

Watch out Common mistakes

Wrong signs with $\sin$ and $\cos$ . $\displaystyle\int\sin x\,dx=-\cos x+C$ , while $\displaystyle\int\cos x\,dx=\sin x+C$ .
Using the power rule for $\dfrac1x$ or $e^x$ . $\dfrac1x=x^{-1}$ is the case $\alpha=-1$ , and its antiderivative is $\ln\lvert x\rvert$ .
Dropping the modulus when finding an area. Wrong: the area between $y=x^3-x$ and the $x$ -axis on $[0, 2]$ is $\displaystyle\int_0^2(x^3-x)\,dx=2$ . Right: split at $x=1$ to get $\dfrac14+\dfrac94=\dfrac52$ .
Forgetting to square, or forgetting $\pi$ , in the volume formula. $V=\pi\displaystyle\int_a^b f^2(x)\,dx$ .
Forgetting that the integral of a product is not the product of the integrals.

Key terms

English	Tiếng Việt
antiderivative (indefinite integral)	nguyên hàm
family of antiderivatives	họ nguyên hàm
definite integral	tích phân
upper / lower limit	cận trên / cận dưới
fundamental theorem of calculus (Newton–Leibniz formula)	công thức Newton – Leibniz
plane region	hình phẳng
solid of revolution	khối tròn xoay
distance / velocity	quãng đường / vận tốc

Exercises

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

Exercise 1 Multiple choice

$\displaystyle\int\left(4x^3-\frac{1}{x^2}\right)dx$ equals A. $x^4+\dfrac1x+C$ . B. $x^4-\dfrac1x+C$ . C. $12x^2+\dfrac{2}{x^3}+C$ . D. $x^4+\dfrac{1}{3x^3}+C$ .

What you need and how to start

What you need: $\dfrac{1}{x^2}=x^{-2}$ and $\displaystyle\int x^{\alpha}\,dx=\dfrac{x^{\alpha+1}}{\alpha+1}+C$ .

Steps:

$\displaystyle\int x^{-2}\,dx=\dfrac{x^{-1}}{-1}=-\dfrac1x$ .
Watch the minus sign in front.

Show solution

$\displaystyle\int4x^3\,dx=x^4$ and $-\displaystyle\int x^{-2}\,dx=-\left(-\dfrac1x\right)=\dfrac1x$ . Option B has the wrong sign; C is the derivative; D lowers the power instead of raising it.

Answer. A.

Exercise 2 Multiple choice

$\displaystyle\int_0^{\frac{\pi}{2}}\cos x\,dx$ equals A. $1$ . B. $0$ . C. $-1$ . D. $\dfrac{\pi}{2}$ .

What you need and how to start

What you need: an antiderivative of $\cos x$ is $\sin x$ .

Steps:

Work out $\sin\dfrac{\pi}{2}-\sin0$ .

Show solution

$\Big[\sin x\Big]_0^{\frac{\pi}{2}}=1-0=1$ . Option C uses the wrong antiderivative $-\sin x$ .

Answer. A.

Exercise 3 Short answer

Evaluate $\displaystyle\int_0^2\left(3x^2-2x+1\right)dx$ .

What you need and how to start

What you need: the Newton–Leibniz formula.

Steps:

An antiderivative is $x^3-x^2+x$ .
Substitute the limits $2$ and $0$ .

Show solution

$\Big[x^3-x^2+x\Big]_0^2=8-4+2=6$ .

Answer. $6$ .

Exercise 4 True/False

Let $f(x)=e^x+2x$ .

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

b) $\displaystyle\int_0^1 f(x)\,dx=e+1$ .

c) The antiderivative $F$ of $f$ with $F(0)=3$ is $F(x)=e^x+x^2+3$ .

d) $\displaystyle\int_{-1}^1 f(x)\,dx=e-\dfrac1e$ .

What you need and how to start

What you need: check an antiderivative by differentiating; $e^0=1$ .

Steps:

Write $F(x)=e^x+x^2+C$ .
Evaluate each integral with $\Big[e^x+x^2\Big]$ .

Show solution

a) True: $(e^x+x^2)'=e^x+2x$ .

b) False: $\Big[e^x+x^2\Big]_0^1=(e+1)-(1+0)=e$ .

c) False: $F(0)=1+C=3$ , so $C=2$ and $F(x)=e^x+x^2+2$ .

d) True: $(e+1)-\left(\dfrac1e+1\right)=e-\dfrac1e$ .

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

Exercise 5 Short answer

Find the area of the region bounded by the parabola $y=x^2-4$ and the $x$ -axis. Give your answer to 1 decimal place.

What you need and how to start

What you need: $S=\displaystyle\int_a^b\lvert f(x)\rvert\,dx$ , where $a$ , $b$ are the intersections with the $x$ -axis.

Steps:

Solve $x^2-4=0$ .
On $[-2, 2]$ , $x^2-4\le0$ , so $\lvert x^2-4\rvert=4-x^2$ .

Show solution

$S=\int_{-2}^2(4-x^2)\,dx=\Big[4x-\frac{x^3}{3}\Big]_{-2}^2=\left(8-\frac83\right)-\left(-8+\frac83\right)=\frac{32}{3}\approx10.67.$

Answer. $10.7$ .

Exercise 6 Written answer

Find the area of the region bounded by the graph of $y=x^3-x$ , the $x$ -axis and the lines $x=0$ , $x=2$ .

What you need and how to start

What you need: split the interval at the roots of $f(x)$ to remove the modulus.

Steps:

$x^3-x=x(x-1)(x+1)$ : negative on $(0, 1)$ , positive on $(1, 2)$ .
Evaluate the two integrals and add.

Show solution

$S=\int_0^1(x-x^3)\,dx+\int_1^2(x^3-x)\,dx=\left(\frac12-\frac14\right)+\left[\left(4-2\right)-\left(\frac14-\frac12\right)\right]=\frac14+\frac94=\frac52.$

Answer. $S=\dfrac52$ .

Exercise 7 Multiple choice

Rotating the region bounded by $y=2x$ , the $x$ -axis and the line $x=3$ about the $x$ -axis gives a cone of volume A. $36\pi$ . B. $9\pi$ . C. $108\pi$ . D. $18\pi$ .

What you need and how to start

What you need: $V=\pi\displaystyle\int_a^b f^2(x)\,dx$ , here with $f^2(x)=4x^2$ .

Steps:

Work out $\pi\displaystyle\int_0^3 4x^2\,dx$ .
Check with the cone formula $\dfrac13\pi r^2h$ .

Show solution

$V=\pi\Big[\dfrac{4x^3}{3}\Big]_0^3=36\pi$ . Check: the cone has $r=6$ , $h=3$ , so $V=\dfrac13\pi\cdot36\cdot3=36\pi$ . Option B forgets to square; C forgets to divide by $3$ ; D squares only $x$ .

Answer. A.

Exercise 8 Short answer

Water flows into a tank at a rate of $r(t)=0.6t+2$ (litres per minute), with $t$ in minutes. How many litres flow into the tank in the first $10$ minutes?

What you need and how to start

What you need: the amount of water is $\displaystyle\int_0^{10} r(t)\,dt$ .

Steps:

An antiderivative is $0.3t^2+2t$ .
Substitute the limits.

Show solution

$\Big[0.3t^2+2t\Big]_0^{10}=30+20=50$ .

Answer. $50$ .

Exercise 9 True/False

A car brakes and slows down uniformly with velocity $v(t)=12-3t$ (m/s), where $t$ is the time in seconds since braking.

a) The car stops after $4$ seconds.

b) The distance travelled $t$ seconds after braking is $s(t)=12t-1.5t^2$ (m), $0\le t\le4$ .

c) From braking to stopping, the car travels $24$ m.

d) In the last second before stopping, the car travels $3$ m.

What you need and how to start

What you need: $s(t)=\displaystyle\int_0^t v(u)\,du$ ; the distance from $t=a$ to $t=b$ is $s(b)-s(a)$ .

Steps:

Solve $v(t)=0$ .
Work out $s(4)$ and $s(4)-s(3)$ .

Show solution

a) True: $12-3t=0\iff t=4$ .

b) True: $s(t)=\Big[12u-1.5u^2\Big]_0^t=12t-1.5t^2$ .

c) True: $s(4)=48-24=24$ .

d) False: $s(4)-s(3)=24-22.5=1.5$ m.

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

Exercise 10 Challenge

A gateway has the shape of a parabola: it is $8$ m wide at ground level and its top is $4$ m above the ground. The whole gateway is filled with glass. Find the area of glass, in m², to 1 decimal place.

What you need and how to start

What you need: put the origin at the middle of the base with the $y$ -axis through the top; the area is $\displaystyle\int_{-4}^4 f(x)\,dx$ .

Steps:

The parabola passes through $(-4, 0)$ , $(4, 0)$ , $(0, 4)$ : $y=4-\dfrac{x^2}{4}$ .
Evaluate the integral.

Show solution

The parabola has the form $y=4-ax^2$ and passes through $(4, 0)$ , so $a=\dfrac14$ . $S=\int_{-4}^4\left(4-\frac{x^2}{4}\right)dx=\Big[4x-\frac{x^3}{12}\Big]_{-4}^4=2\left(16-\frac{16}{3}\right)=\frac{64}{3}\approx21.33.$

Answer. $21.3$ .

Antiderivatives

Definite integrals

Area and volume

Real-life applications

Key terms

Exercises

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) The area of $H$ is $S=\displaystyle\int_0^3x^2\,dx$ .	True	False
b) $S=9$ .	True	False
c) The volume of the solid formed by rotating $H$ about the $x$ -axis is $V=\pi\displaystyle\int_0^3x^2\,dx$ .	True	False
d) $V=\dfrac{243\pi}{5}$ .	True	False