Definite integrals

A motorbike accelerates for $3$ seconds. Its velocity after $t$ seconds is $v(t)=t^2+2$ (m/s): $2$ m/s at the start, $3$ m/s after $1$ second, $6$ m/s after $2$ seconds and $11$ m/s after $3$ seconds. How many metres does it travel in those $3$ seconds?

A quick guess is to average the first and last velocities: $\dfrac{2+11}{2}\cdot3=19.5$ m. That only works when the velocity rises steadily, but here it rises faster and faster. Let us find the right answer.

The area under a graph

Explore: cutting time into small pieces

If the velocity is constant, the distance is velocity times time, which is the area of a rectangle. When the velocity changes, we cut $[0, 3]$ into small pieces and treat the velocity as almost constant on each one.

With pieces $1$ second long, taking the velocity at the start of each piece gives an underestimate, and taking it at the end gives an overestimate:

$1\cdot(2+3+6)=11,\qquad 1\cdot(3+6+11)=20.$

Length of each piece	$1$	$0.5$	$0.1$
Underestimate	$11$	$12.875$	$14.555$
Overestimate	$20$	$17.375$	$15.455$

The two estimates close in on one number, $15$ . This number is the area of the region under the graph of $v(t)$ from $t=0$ to $t=3$ .

Distance as the area under

v(t)=t^2+2

Here is the surprise: if $S(t)=\dfrac{t^3}{3}+2t$ is an antiderivative of $v(t)$ , then $S(3)-S(0)=9+6-0=15$ , exactly the same number. This is no coincidence.

The definition of the integral

The result does not depend on which antiderivative we choose: replacing $F$ by $F+C$ gives $(F(b)+C)-(F(a)+C)=F(b)-F(a)$ . Nor does it depend on the name of the variable: $\displaystyle\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$ .

Why is it true? Let $S(x)$ be the area under the graph from $a$ to $x$ . When $x$ grows by a small amount $h$ , the area grows by a thin strip that is almost a rectangle, so $S(x+h)-S(x)\approx f(x)\cdot h$ . Dividing by $h$ and letting $h\to0$ gives $S'(x)=f(x)$ , so $S$ is an antiderivative of $f$ with $S(a)=0$ . The area we want is therefore $S(b)=S(b)-S(a)=F(b)-F(a)$ .

The value at the lower limit is not always $0$ , especially for trigonometric functions.

Properties of the integral

The first two come from the rules for antiderivatives. The third holds because $\big(F(c)-F(a)\big)+\big(F(b)-F(c)\big)=F(b)-F(a)$ ; geometrically, we cut a region into two pieces.

Example Using the properties without knowing the function

Given $\displaystyle\int_0^2 f(x)\,dx=3$ , $\displaystyle\int_2^5 f(x)\,dx=-1$ and $\displaystyle\int_0^5 g(x)\,dx=4$ , find $\displaystyle\int_0^5\big(2f(x)-3g(x)\big)\,dx$ and $\displaystyle\int_5^0 f(x)\,dx$ .

Solution. We join the adjacent intervals $[0, 2]$ and $[2, 5]$ :

$\int_0^5 f(x)\,dx=3+(-1)=2.$

We split the sum and take the constants outside:

$\int_0^5\big(2f(x)-3g(x)\big)\,dx=2\cdot2-3\cdot4=-8.$

Swapping the limits changes the sign: $\displaystyle\int_5^0 f(x)\,dx=-2$ .

With an absolute value, we find where the expression changes sign and split the interval using the third property.

Example An integral with an absolute value

Evaluate $\displaystyle\int_0^2\lvert x^2-1\rvert\,dx$ .

Solution. On $[0, 2]$ , $x^2-1$ changes sign at $x=1$ : it is negative on $[0, 1)$ and positive on $(1, 2]$ . We remove the absolute value on each piece:

$\begin{aligned} \int_0^2\lvert x^2-1\rvert\,dx&=\int_0^1(1-x^2)\,dx+\int_1^2(x^2-1)\,dx\\ &=\Big[x-\frac{x^3}{3}\Big]_0^1+\Big[\frac{x^3}{3}-x\Big]_1^2=\frac23+\left(\frac23+\frac23\right)=2. \end{aligned}$

So the integral equals $2$ , the total area of the two shaded parts in the figure below.

The graph of

y=\lvert x^2-1\rvert

[0, 2]

Functions of the form $f(ax+b)$

If $F$ is an antiderivative of $f$ , the chain rule (grade 11) gives $\big(F(ax+b)\big)'=a\,f(ax+b)$ . Dividing by $a$ gives the rule:

$\int f(ax+b)\,dx=\frac1a F(ax+b)+C\qquad(a\neq0).$

For example, $\displaystyle\int\cos 2x\,dx=\frac12\sin2x+C$ and $\displaystyle\int e^{3x}dx=\frac13e^{3x}+C$ . This is the simplest case of integration by substitution, with $u=ax+b$ .

Back to the motorbike

Watch out Common mistakes

Subtracting the wrong way round. Wrong: $\displaystyle\int_1^2(3x^2-2x+1)\,dx=F(1)-F(2)=-5$ . Why it is wrong: the definition is $F(\text{upper limit})-F(\text{lower limit})$ . Right: $F(2)-F(1)=5$ .
Taking the value at the lower limit as $0$ . Wrong: $\displaystyle\int_0^{\frac{\pi}{2}}\sin x\,dx=-\cos\frac{\pi}{2}=0$ . Why it is wrong: $-\cos0=-1\neq0$ . Right: $\Big[-\cos x\Big]_0^{\frac{\pi}{2}}=0-(-1)=1$ .
Moving the absolute value outside. Wrong: $\displaystyle\int_0^2\lvert x^2-1\rvert\,dx=\left\lvert\int_0^2(x^2-1)\,dx\right\rvert=\frac23$ . Why it is wrong: the negative and positive parts cancel inside the inner integral. Right: split at $x=1$ to get $2$ .
Forgetting the factor $\dfrac1a$ . Wrong: $\displaystyle\int_0^1(2x+1)^3dx=\Big[\frac{(2x+1)^4}{4}\Big]_0^1=20$ . Why it is wrong: the derivative of $\dfrac{(2x+1)^4}{4}$ is $2(2x+1)^3$ . Right: divide by $2$ as well to get $10$ .

Key terms

English	Tiếng Việt
definite integral	tích phân
lower limit, upper limit	cận dưới, cận trên
fundamental theorem of calculus	công thức Newton – Leibniz
integrand	hàm dưới dấu tích phân
area under the graph	diện tích dưới đồ thị
approximating sum (Riemann sum)	tổng xấp xỉ diện tích
substitution	đổi biến
absolute value	giá trị tuyệt đối

Summary

$\displaystyle\int_a^b f(x)\,dx=\Big[F(x)\Big]_a^b=F(b)-F(a)$ , where $F$ is any antiderivative of $f$ .
If $f\ge0$ , the integral is the area under the graph; if the velocity is positive, the integral is the distance travelled.
$\displaystyle\int_a^a=0$ , $\displaystyle\int_b^a=-\int_a^b$ , $\displaystyle\int_a^b=\int_a^c+\int_c^b$ ; constant factors and sums split apart.
With an absolute value: find where the sign changes, split the interval, remove the bars on each piece.
$\displaystyle\int f(ax+b)\,dx=\frac1aF(ax+b)+C$ .

Practice

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

Exercise 1

Evaluate $\displaystyle\int_0^2(x^3-2x+3)\,dx$ .

What you need and how to start

What you need: $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$ .

Steps:

Find an antiderivative $F(x)=\dfrac{x^4}{4}-x^2+3x$ .
Compute $F(2)-F(0)$ .

Show solution

$\Big[\dfrac{x^4}{4}-x^2+3x\Big]_0^2=(4-4+6)-0=6$ .

Answer. $6$ .

Exercise 2

Evaluate $\displaystyle\int_1^4\left(\sqrt x-\frac{1}{x^2}\right)dx$ .

What you need and how to start

What you need: $\sqrt x=x^{\frac12}$ has antiderivative $\dfrac23x\sqrt x$ ; $x^{-2}$ has antiderivative $-\dfrac1x$ .

Steps:

Write an antiderivative $F(x)=\dfrac23x\sqrt x+\dfrac1x$ .
Compute $F(4)-F(1)$ , taking care with the fractions.

Show solution

$F(4)=\dfrac{16}{3}+\dfrac14$ and $F(1)=\dfrac23+1$ . The difference is $\dfrac{14}{3}-\dfrac34=\dfrac{56-9}{12}=\dfrac{47}{12}$ .

Answer. $\dfrac{47}{12}$ .

Exercise 3

Evaluate $\displaystyle\int_0^{\pi}(\sin x+2)\,dx$ .

What you need and how to start

What you need: the antiderivative of $\sin x$ is $-\cos x$ ; $\cos\pi=-1$ and $\cos0=1$ .

Steps:

Write $F(x)=-\cos x+2x$ .
Compute $F(\pi)-F(0)$ .

Show solution

$F(\pi)=1+2\pi$ and $F(0)=-1$ . So the integral equals $1+2\pi-(-1)=2+2\pi$ .

Answer. $2+2\pi$ .

Exercise 4

Given $\displaystyle\int_1^3 f(x)\,dx=5$ and $\displaystyle\int_1^3 g(x)\,dx=-2$ , find $\displaystyle\int_1^3\big(3f(x)-2g(x)+x\big)\,dx$ .

What you need and how to start

What you need: split the sum and take constants outside the integral.

Steps:

Write it as $3\cdot5-2\cdot(-2)+\displaystyle\int_1^3x\,dx$ .
Evaluate $\displaystyle\int_1^3x\,dx$ .

Show solution

$\displaystyle\int_1^3x\,dx=\Big[\frac{x^2}{2}\Big]_1^3=\frac92-\frac12=4$ . The total is $15+4+4=23$ .

Answer. $23$ .

Exercise 5

Consider the integral $I=\displaystyle\int_0^3\lvert x-2\rvert\,dx$ . Is each claim true or false?

a) On $[0, 2]$ , $\lvert x-2\rvert=2-x$ .

b) $I=\left\lvert\displaystyle\int_0^3(x-2)\,dx\right\rvert=\dfrac32$ .

c) $I=\dfrac52$ .

d) $\displaystyle\int_3^0\lvert x-2\rvert\,dx=\dfrac52$ .

What you need and how to start

What you need: split the interval where $x-2$ changes sign; swapping the limits changes the sign.

Steps:

Find the sign of $x-2$ on $[0, 2]$ and on $[2, 3]$ .
Evaluate the two smaller integrals and add.

Show solution

a) True: for $x\le2$ , $x-2\le0$ .

b) False: the absolute value cannot be moved outside; $I$ is not $\dfrac32$ (see c).

c) True: $\displaystyle\int_0^2(2-x)\,dx=2$ and $\displaystyle\int_2^3(x-2)\,dx=\frac12$ , so $I=\dfrac52$ .

d) False: swapping the limits changes the sign, so that integral equals $-\dfrac52$ .

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

Exercise 6

Evaluate $\displaystyle\int_0^1(3x+1)^2\,dx$ .

What you need and how to start

What you need: $\displaystyle\int f(ax+b)\,dx=\frac1aF(ax+b)+C$ , here with $a=3$ .

Steps:

An antiderivative is $\dfrac{(3x+1)^3}{9}$ .
Substitute the limits $1$ and $0$ .

Show solution

$\Big[\dfrac{(3x+1)^3}{9}\Big]_0^1=\dfrac{64}{9}-\dfrac19=7$ .

Answer. $7$ .

Exercise 7

Water flows into a tank at the rate $r(t)=40-2t$ (litres per minute), $0\le t\le10$ . How much water flows into the tank in the first $10$ minutes?

What you need and how to start

What you need: the amount that flows in from $t=a$ to $t=b$ is $\displaystyle\int_a^b r(t)\,dt$ (the rate is non-negative).

Steps:

Write $\displaystyle\int_0^{10}(40-2t)\,dt$ .
Evaluate it with the antiderivative $40t-t^2$ .

Show solution

$\Big[40t-t^2\Big]_0^{10}=400-100=300$ .

Answer. $300$ litres.

Exercise 8 Review

$F(x)$ is an antiderivative of $f(x)$ on $\mathbb{R}$ with $F(1)=2$ and $F(4)=7$ . Find $\displaystyle\int_1^4 f(x)\,dx$ and $\displaystyle\int_1^4\big(2f(x)+1\big)\,dx$ .

What you need and how to start

What you need: the definition of the integral through an antiderivative (see antiderivatives).

Steps:

$\displaystyle\int_1^4f(x)\,dx=F(4)-F(1)$ .
Split the sum; $\displaystyle\int_1^41\,dx=3$ .

Show solution

$\displaystyle\int_1^4f(x)\,dx=7-2=5$ . Then $\displaystyle\int_1^4\big(2f(x)+1\big)\,dx=2\cdot5+3=13$ .

Answer. $5$ and $13$ .

Exercise 9 Review

Evaluate $\displaystyle\int_0^{\ln3}e^x\,dx$ and $\displaystyle\int_1^2 2^x\,dx$ (give the second one to 2 decimal places).

What you need and how to start

What you need: $e^{\ln a}=a$ (logarithms, grade 11); the antiderivative of $2^x$ is $\dfrac{2^x}{\ln2}$ .

Steps:

$\Big[e^x\Big]_0^{\ln3}=e^{\ln3}-e^0$ .
$\Big[\dfrac{2^x}{\ln2}\Big]_1^2=\dfrac{4-2}{\ln2}$ .

Show solution

$\displaystyle\int_0^{\ln3}e^x\,dx=3-1=2$ . $\displaystyle\int_1^22^x\,dx=\frac{2}{\ln2}\approx2.885$ .

Answer. $2$ and $\dfrac{2}{\ln2}\approx2.89$ .

Exercise 10 Challenge

Evaluate $\displaystyle\int_{-1}^2\lvert x^2-x\rvert\,dx$ .

What you need and how to start

What you need: $x^2-x=x(x-1)$ changes sign at $0$ and $1$ ; it is positive outside the roots and negative between them.

Steps:

Split into the three intervals $[-1, 0]$ , $[0, 1]$ , $[1, 2]$ .
Use the antiderivative $G(x)=\dfrac{x^3}{3}-\dfrac{x^2}{2}$ on each piece, changing the sign on the middle one.

Show solution

$G(-1)=-\dfrac56$ , $G(0)=0$ , $G(1)=-\dfrac16$ , $G(2)=\dfrac23$ . The three parts: $G(0)-G(-1)=\dfrac56$ ; $-(G(1)-G(0))=\dfrac16$ ; $G(2)-G(1)=\dfrac56$ . The total is $\dfrac56+\dfrac16+\dfrac56=\dfrac{11}{6}$ .

Answer. $\dfrac{11}{6}$ .

The area under a graph

Explore: cutting time into small pieces

The definition of the integral

Properties of the integral

Functions of the form $f(ax+b)$

Back to the motorbike

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $\displaystyle\int_0^3 f(x)\,dx=0$ .	True	False
b) $f(x)\le0$ for every $x\in[1, 3]$ .	True	False
c) $\displaystyle\int_0^3\lvert f(x)\rvert\,dx=0$ .	True	False
d) $\displaystyle\int_0^3\lvert f(x)\rvert\,dx=\dfrac83$ .	True	False

Definite integrals

The area under a graph

Explore: cutting time into small pieces

The definition of the integral

Properties of the integral

Functions of the form f(ax+b)f(ax+b)f(ax+b)

Back to the motorbike

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Functions of the form $f(ax+b)$