Learn9709Paper 1 · Pure Mathematics 1IntegrationDefinite integrals

Integration · Subtopic 3 of 5

Definite Integrals

What $\displaystyle\int_a^b f(x)\,dx$ means and how to evaluate it, why there is no $+c$ , how to find an unknown limit, and how some integrals with an infinite limit still give a finite answer.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Evaluate $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)$ with square-bracket notation.
Explain why the constant of integration cancels.
Find an unknown limit from the value of a definite integral.
Evaluate simple improper integrals with an infinite limit or an integrand undefined at a limit.

Pause and think

A car's velocity is $v=3t^2$ m/s. Its position is $s=t^3+c$ . How far does it travel between $t=1$ and $t=3$ ? Does the value of $c$ matter?

Section 2 of 14

Core idea

Start hereDistance from a curved speed graph

If you drive at a steady $20$ m/s for $10$ s, you go $200$ m: the area of a rectangle under the speed graph. But if your speed is $v=t^2$ , the graph curves. Chop the time into thin strips, each almost a rectangle, and add them. As the strips get thinner, the total becomes exact, and astonishingly, it's given by the antiderivative. This is the Fundamental Theorem of Calculus.

The big idea

$\displaystyle\int_a^bf(x)\,dx=F(b)-F(a)$ , where $F$ is any antiderivative of $f$ . When $f\ge0$ this is the area under $y=f(x)$ from $x=a$ to $x=b$ .

1. Area as a limit of strips

The area under y = x² from 1 to 3

upper limit = 1.05

As the upper limit moves from

1

3

, the shaded area grows from

0

\dfrac{27-1}{3}=\dfrac{26}{3}

Why antiderivatives give areas

$A(x)=\text{area under }f\text{ from }a\text{ to }x$
An "area so far" function.
$A(x+h)-A(x)\approx f(x)\cdot h$
The extra thin strip is nearly a rectangle of height $f(x)$ , width $h$ .
$\frac{A(x+h)-A(x)}{h}\approx f(x)\ \Rightarrow\ A'(x)=f(x)$
Divide by $h$ ; let $h\to0$ (definition of derivative).
$A(x)=F(x)+C,\ A(a)=0\Rightarrow C=-F(a)$
$A$ is an antiderivative; no area from $a$ to $a$ .
$\text{Area}=A(b)=F(b)-F(a)$
Put $x=b$ .

\int_a^bf(x)\,dx=\Big[F(x)\Big]_a^b=F(b)-F(a)

$a$ , $b$: the lower and upper limits.
$[F(x)]_a^b$: shorthand: evaluate at b, subtract the value at a.

2. Evaluating

Worked example $\displaystyle\int_1^3x^2\,dx$

Evaluate.

Goal: A number.
Strategy: Integrate (no $+c$ needed), top minus bottom.

$\left[\frac{x^3}3\right]_1^3$
Antiderivative; any $c$ would cancel.
$=\frac{27}3-\frac13=\frac{26}3\approx8.67$
Upper minus lower.

Answer

\frac{26}3

✓ CheckThe region sits in a

2

9

box (area

18

) and above a

2

1

box:

2<8.67<18

✓.

Worked example $\displaystyle\int_1^4\left(3\sqrt x-\frac2{x^2}\right)dx$

Evaluate.

Goal: A number.
Strategy: Powers, integrate, substitute.

$\left[2x^{3/2}+\frac2x\right]_1^4$
$\int3x^{1/2}=2x^{3/2}$ ; $\int-2x^{-2}=2x^{-1}$ .
$=\left(16+\tfrac12\right)-(2+2)=12.5$
$4^{3/2}=8$ .

Answer

12.5

✓ CheckDifferentiate

2x^{3/2}+\frac2x

3x^{1/2}-\frac2{x^2}

✓.

3. Improper integrals

Worked example $\displaystyle\int_1^\infty\frac1{x^2}\,dx$

Evaluate.

Goal: A finite value, if it exists.
Strategy: Use a finite upper limit $N$ , then let $N\to\infty$ .

$\left[-\frac1x\right]_1^N=1-\frac1N$
Integrate.
$N\to\infty:\ \frac1N\to0$
Limit.

Answer

1

: an infinitely long region with finite area.

✓ CheckBy contrast

\int_1^N\frac1x\,dx=\ln N\to\infty

: that one doesn't converge.

Pause and think

Find $k>1$ with $\int_1^k(2x-1)\,dx=6$ .

Summary

$\int_a^bf=F(b)-F(a)$ .
Area = limit of strips (Fundamental Theorem).
Infinite limits: use $N$ then $N\to\infty$ .

Definite integral: Integral with limits; a number.
Improper integral: Integral with an infinite limit.

The definite integral as an area: ∫ from 1 to 3 of x² dx

\displaystyle\int_1^3 x^2\,dx=\left[\tfrac{x^3}{3}\right]_1^3=9-\tfrac13=\tfrac{26}{3}

: the shaded area between the curve, the

x

-axis and the lines

x=1

and

x=3

HistoryThe Fundamental Theorem. That the area under a curve can be found by reversing differentiation is called the Fundamental Theorem of Calculus. Isaac Barrow, Newton's teacher at Cambridge, came close to it in the 1660s; Newton and Leibniz turned it into a working method. It links the two halves of calculus.

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