Topics

Definite integrals

Covers AS91579

What a definite integral is

A definite integral has limits — a start value and an end value — and evaluates to a number, not a family of functions.

$\int_a^b f(x)\,dx$

Each part:
- $a$ — the lower limit, written at the bottom
- $b$ — the upper limit, at the top
- $f(x)$ — the integrand
There is no $+c$ . The constants cancel when you subtract, which is why a definite integral produces a definite number.

Evaluating one

$\int_a^b f(x)\,dx = \left[F(x)\right]_a^b = F(b) - F(a)$

where $F$ is any anti-derivative of $f$ .

The procedure:
- Integrate, and write the result in square brackets with the limits attached.
- Substitute the upper limit into the anti-derivative.
- Substitute the lower limit.
- Subtract: upper minus lower.
Worked through:

$\int_1^3 \left(2x + 1\right)dx = \left[x^2 + x\right]_1^3$

$= \left(9 + 3\right) - \left(1 + 1\right) = 12 - 2 = 10$

Note the bracket discipline. $F(b) - F(a)$ needs brackets around the whole of $F(a)$ , or the minus sign will only apply to its first term. This is the most frequent arithmetic error in the topic.

Why the constant disappears

If $F(x) + c$ is used instead of $F(x)$ :

$\left[F(b) + c\right] - \left[F(a) + c\right] = F(b) - F(a)$

The $c$ cancels, so it never needs to be written for a definite integral.

Properties worth knowing

Swapping the limits changes the sign:

$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

Equal limits give zero:

$\int_a^a f(x)\,dx = 0$

Adjacent intervals join:

$\int_a^b f(x)\,dx + \int_b^d f(x)\,dx = \int_a^d f(x)\,dx$

Constants factor out, and sums split, exactly as for indefinite integrals.

The link with area

Geometrically, $\displaystyle\int_a^b f(x)\,dx$ $\int_{a}^{b} f (x) d x$ is the signed area between the curve and the $x$ $x$ -axis:
- Where the curve is above the axis, the contribution is positive.
- Where it is below, the contribution is negative.
So a definite integral can be zero even when there is plenty of area, if the positive and negative parts cancel.
"Evaluate the integral" and "find the area" are different instructions. The next page deals with that difference properly.

Definite integrals with an unknown limit

A common exam form: "find $k$ $k$ such that $\displaystyle\int_0^k \left(6x - 2\right)dx = 15$ $\int_{0}^{k} (6 x - 2) d x = 15$ ".
- Integrate: $\left[3x^2 - 2x\right]_0^k = 3k^2 - 2k - 0 = 3k^2 - 2k$
- Set equal: $3k^2 - 2k = 15$
- Solve: $3k^2 - 2k - 15 = 0$ , giving $(3k + 7)(k - 3) = 0$ , so $k = 3$ or $k = -\frac{7}{3}$
- Check both against any restriction in the question — if $k > 0$ was stated, only $k = 3$ survives.

Working with radians and exact values

Definite integrals of trigonometric functions almost always use limits that are multiples of $\pi$ .
Keep everything exact through the working; round only at the end, and only if asked.
Worked through:

$\int_0^{\pi/2}\sin x\,dx = \left[-\cos x\right]_0^{\pi/2} = \left(-\cos\frac{\pi}{2}\right) - \left(-\cos 0\right)$

$= (0) - (-1) = 1$

Tips and Tricks

Brackets around both substitutions. Write $\left[F(x)\right]_a^b = \left(F(b)\right) - \left(F(a)\right)$ with the brackets visible. The 2025 report lists "used brackets carefully and reliably" as a Merit behaviour, which tells you how often they are not.

Upper minus lower, in that order. Reversing it flips the sign of your answer.

No $+c$ on a definite integral. Including it is not wrong mathematically — it cancels — but it wastes time and suggests you have not understood why.

The report notes candidates who "integrated and substituted correctly, but then were not successful in completing the problem because of algebraic errors". The integration is often the easy part; slow down on the arithmetic that follows.

Keep full accuracy through the substitutions and round only at the very end. The specification says rounding errors are not penalised, but rounding early produces genuinely wrong answers, which are.

If a question asks for $k$ and your equation is quadratic, expect two answers — then check each against the context before discarding either.

Worked ExampleEvaluating and solving for a limit

(a) Evaluate $\displaystyle\int_0^2 \left(3x^2 - 4x + 1\right)dx$ . (b) Find the positive value of $k$ for which $\displaystyle\int_1^k \frac{4}{x^2}\,dx = 3$ .

Part (a), Step 1 — Integrate

Apply the power rule to each term and write the result in square brackets with the limits attached:

$\int_0^2 \left(3x^2 - 4x + 1\right)dx = \left[x^3 - 2x^2 + x\right]_0^2$

Part (a), Step 2 — Substitute the upper limit

$F(2) = 8 - 2(4) + 2 = 8 - 8 + 2 = 2$

Part (a), Step 3 — Substitute the lower limit

$F(0) = 0 - 0 + 0 = 0$

Part (a), Step 4 — Subtract, upper minus lower

$\int_0^2 \left(3x^2 - 4x + 1\right)dx = (2) - (0) = 2$

Part (b), Step 1 — Rewrite in index form

A root or reciprocal must be converted before integrating:

$\int_1^k 4x^{-2}\,dx$

Part (b), Step 2 — Integrate

Raise the power to $-1$ and divide by $-1$ :

$= \left[\frac{4x^{-1}}{-1}\right]_1^k = \left[-\frac{4}{x}\right]_1^k$

Part (b), Step 3 — Substitute both limits

Upper minus lower, with brackets around each:

$= \left(-\frac{4}{k}\right) - \left(-\frac{4}{1}\right)$

$= -\frac{4}{k} + 4$

Part (b), Step 4 — Set equal to the given value

$4 - \frac{4}{k} = 3$

Part (b), Step 5 — Solve

$-\frac{4}{k} = -1$

$\frac{4}{k} = 1 \quad \Rightarrow \quad k = 4$

Part (b), Step 6 — Check

$\int_1^4 4x^{-2}dx = \left[-\frac{4}{x}\right]_1^4 = \left(-1\right) - \left(-4\right) = 3 \;\checkmark$

Also check the limit is valid: the integrand $\frac{4}{x^2}$ is undefined at $x = 0$ , but the interval $1 \leq x \leq 4$ avoids it entirely ✓

Definite integrals

Covers AS91579

What a definite integral is

A definite integral has limits — a start value and an end value — and evaluates to a number, not a family of functions.

$\int_a^b f(x)\,dx$

Each part:
- $a$ — the lower limit, written at the bottom
- $b$ — the upper limit, at the top
- $f(x)$ — the integrand
There is no $+c$ . The constants cancel when you subtract, which is why a definite integral produces a definite number.

Evaluating one

$\int_a^b f(x)\,dx = \left[F(x)\right]_a^b = F(b) - F(a)$

where $F$ is any anti-derivative of $f$ .

The procedure:
- Integrate, and write the result in square brackets with the limits attached.
- Substitute the upper limit into the anti-derivative.
- Substitute the lower limit.
- Subtract: upper minus lower.
Worked through:

$\int_1^3 \left(2x + 1\right)dx = \left[x^2 + x\right]_1^3$

$= \left(9 + 3\right) - \left(1 + 1\right) = 12 - 2 = 10$

Note the bracket discipline. $F(b) - F(a)$ needs brackets around the whole of $F(a)$ , or the minus sign will only apply to its first term. This is the most frequent arithmetic error in the topic.

Why the constant disappears

If $F(x) + c$ is used instead of $F(x)$ :

$\left[F(b) + c\right] - \left[F(a) + c\right] = F(b) - F(a)$

The $c$ cancels, so it never needs to be written for a definite integral.

Properties worth knowing

Swapping the limits changes the sign:

$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$

Equal limits give zero:

$\int_a^a f(x)\,dx = 0$

Adjacent intervals join:

$\int_a^b f(x)\,dx + \int_b^d f(x)\,dx = \int_a^d f(x)\,dx$

Constants factor out, and sums split, exactly as for indefinite integrals.

The link with area

Geometrically, $\displaystyle\int_a^b f(x)\,dx$ $\int_{a}^{b} f (x) d x$ is the signed area between the curve and the $x$ $x$ -axis:
- Where the curve is above the axis, the contribution is positive.
- Where it is below, the contribution is negative.
So a definite integral can be zero even when there is plenty of area, if the positive and negative parts cancel.
"Evaluate the integral" and "find the area" are different instructions. The next page deals with that difference properly.

Definite integrals with an unknown limit

A common exam form: "find $k$ $k$ such that $\displaystyle\int_0^k \left(6x - 2\right)dx = 15$ $\int_{0}^{k} (6 x - 2) d x = 15$ ".
- Integrate: $\left[3x^2 - 2x\right]_0^k = 3k^2 - 2k - 0 = 3k^2 - 2k$
- Set equal: $3k^2 - 2k = 15$
- Solve: $3k^2 - 2k - 15 = 0$ , giving $(3k + 7)(k - 3) = 0$ , so $k = 3$ or $k = -\frac{7}{3}$
- Check both against any restriction in the question — if $k > 0$ was stated, only $k = 3$ survives.

Working with radians and exact values

Definite integrals of trigonometric functions almost always use limits that are multiples of $\pi$ .
Keep everything exact through the working; round only at the end, and only if asked.
Worked through:

$\int_0^{\pi/2}\sin x\,dx = \left[-\cos x\right]_0^{\pi/2} = \left(-\cos\frac{\pi}{2}\right) - \left(-\cos 0\right)$

$= (0) - (-1) = 1$

Tips and Tricks

Upper minus lower, in that order. Reversing it flips the sign of your answer.

No $+c$ on a definite integral. Including it is not wrong mathematically — it cancels — but it wastes time and suggests you have not understood why.

If a question asks for $k$ and your equation is quadratic, expect two answers — then check each against the context before discarding either.

Worked ExampleEvaluating and solving for a limit

(a) Evaluate $\displaystyle\int_0^2 \left(3x^2 - 4x + 1\right)dx$ . (b) Find the positive value of $k$ for which $\displaystyle\int_1^k \frac{4}{x^2}\,dx = 3$ .

Part (a), Step 1 — Integrate

Apply the power rule to each term and write the result in square brackets with the limits attached:

$\int_0^2 \left(3x^2 - 4x + 1\right)dx = \left[x^3 - 2x^2 + x\right]_0^2$

Part (a), Step 2 — Substitute the upper limit

$F(2) = 8 - 2(4) + 2 = 8 - 8 + 2 = 2$

Part (a), Step 3 — Substitute the lower limit

$F(0) = 0 - 0 + 0 = 0$

Part (a), Step 4 — Subtract, upper minus lower

$\int_0^2 \left(3x^2 - 4x + 1\right)dx = (2) - (0) = 2$

Part (b), Step 1 — Rewrite in index form

A root or reciprocal must be converted before integrating:

$\int_1^k 4x^{-2}\,dx$

Part (b), Step 2 — Integrate

Raise the power to $-1$ and divide by $-1$ :

$= \left[\frac{4x^{-1}}{-1}\right]_1^k = \left[-\frac{4}{x}\right]_1^k$

Part (b), Step 3 — Substitute both limits

Upper minus lower, with brackets around each:

$= \left(-\frac{4}{k}\right) - \left(-\frac{4}{1}\right)$

$= -\frac{4}{k} + 4$

Part (b), Step 4 — Set equal to the given value

$4 - \frac{4}{k} = 3$

Part (b), Step 5 — Solve

$-\frac{4}{k} = -1$

$\frac{4}{k} = 1 \quad \Rightarrow \quad k = 4$

Part (b), Step 6 — Check

$\int_1^4 4x^{-2}dx = \left[-\frac{4}{x}\right]_1^4 = \left(-1\right) - \left(-4\right) = 3 \;\checkmark$

Also check the limit is valid: the integrand $\frac{4}{x^2}$ is undefined at $x = 0$ , but the interval $1 \leq x \leq 4$ avoids it entirely ✓