Year 1 Pure · Chapter 14

Integration

Specification: Edexcel Pure 8 — Integration · AQA section H

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In this chapter

1. The reverse of differentiation
2. Definite integrals and area
3. Finding a curve from its gradient
4. Area between a curve and a line
Practice

Key facts — this chapter

Integration is the reverse of differentiation — it undoes the derivative.
Power rule: $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$ (for $n \neq -1$ ) — raise the power by one, then divide by the new power.
An indefinite integral always carries a constant of integration $+c$ .
A definite integral $\displaystyle\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ — no $+c$ needed.
For $f(x) \geq 0$ , $\displaystyle\int_a^b f(x)\,dx$ is the area under the curve between $x = a$ and $x = b$ .
Given $\dfrac{dy}{dx}$ and one point, integrate and use the point to find $c$ , recovering the curve.

Differentiation finds a gradient from a curve. Integration runs the other way: given the gradient, recover the curve — and, remarkably, the same operation measures the area underneath it.

1. The reverse of differentiation

If differentiating $x^3$ gives $3x^2$ , then integrating $3x^2$ should give back $x^3$ . There’s a catch, though: any constant differentiates to zero, so we can never be sure it wasn’t there — hence the $+c$ .

Key result

\int x^n\,dx = \frac{x^{n+1}}{n+1} + c \qquad (n \neq -1)

Raise the power by one, then divide by the new power — the exact undo of the differentiation rule. The $+c$ is the constant of integration.

A family of parabolas y = x squared + c for several values of c, all the same shape shifted vertically — Integrating 2x gives x² + c — a whole family of curves, one for each c.

Worked example 14.1 · Integrating a polynomial

Find $\displaystyle\int (6x^2 + 2x - 5)\,dx$ .

Integrate term by term, raising each power by one:

\int (6x^2 + 2x - 5)\,dx = \frac{6x^3}{3} + \frac{2x^2}{2} - 5x + c = 2x^3 + x^2 - 5x + c

Worked example 14.2 · Negative and fractional powers

Find $\displaystyle\int \left(\frac{3}{x^2} + \sqrt{x}\right)dx$ .

Rewrite in index form:
$\int \left(3x^{-2} + x^{1/2}\right)dx$
Apply the rule to each (power up by one, divide):
$= \frac{3x^{-1}}{-1} + \frac{x^{3/2}}{3/2} + c = -3x^{-1} + \tfrac{2}{3}x^{3/2} + c$

Common mistake

Never drop the $+c$ on an indefinite integral — it’s part of the answer, and exam mark schemes penalise its absence. (Definite integrals are the exception: the $c$ cancels, so you leave it out there.)

2. Definite integrals and area

Put limits on the integral and it becomes a number — and that number is the area under the curve.

Key result

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

where $F$ is any integral of $f$ . Integrate, substitute the top limit, subtract the bottom limit. For $f(x) \geq 0$ this equals the area between the curve and the $x$ -axis from $a$ to $b$ .

A curve with the region under it shaded between x = a and x = b — For a positive curve, the definite integral from a to b is the shaded area.

Why it works

Slice the area into thin rectangles of width $h$ and height $f(x)$ . Their total is an approximation; as the rectangles get thinner the approximation tightens onto the true area — and that limit is exactly the integral. The same “shrink it to zero” move that built the derivative builds the area.

Two panels: 4 rectangles and 12 rectangles approximating the area under a curve, the 12 hugging the curve much more closely — Thin rectangles approximate the area; in the limit they give the integral.

Approximate the area under y = x² + 1 from x = 0 to 2 with n rectangles. Add more and the total closes in on the exact integral, 14⁄3 ≈ 4.667.

midpoint rectangles · the exact area is the limit as n → ∞

Worked example 14.3 · Evaluating a definite integral

Evaluate $\displaystyle\int_1^3 3x^2\,dx$ .

Integrate (no $+c$ for a definite integral):
$\int 3x^2\,dx = x^3$
Substitute the limits, top minus bottom:
$\big[x^3\big]_1^3 = 3^3 - 1^3 = 27 - 1 = 26$

Worked example 14.4 · Area under a curve

Find the area under $y = x^2 + 1$ between $x = 0$ and $x = 2$ .

Since $y > 0$ here, the area is the definite integral:

\int_0^2 (x^2 + 1)\,dx = \left[\frac{x^3}{3} + x\right]_0^2 = \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3}

3. Finding a curve from its gradient

Because integration recovers a function from its derivative, a known point pins down the otherwise-mysterious $+c$ .

Key result

Given $\dfrac{dy}{dx}$ and a point the curve passes through:

Integrate to get $y$ (with $+c$ );
substitute the known point to find $c$ ;
write the full equation.

Worked example 14.5 · Recovering the curve

A curve passes through $(1, 3)$ and has gradient $\dfrac{dy}{dx} = 6x - 4$ . Find its equation.

Integrate:
$y = \int (6x - 4)\,dx = 3x^2 - 4x + c$
Use $(1, 3)$ to find $c$ :
$3 = 3(1)^2 - 4(1) + c = -1 + c \;\;\Longrightarrow\;\; c = 4$
So the curve is $y = 3x^2 - 4x + 4$ .

4. Area between a curve and a line

To find the area enclosed between two graphs, integrate the difference (top curve minus bottom curve) between their points of intersection.

Key result

The area between $y = f(x)$ (upper) and $y = g(x)$ (lower) from $a$ to $b$ is

\int_a^b \big[f(x) - g(x)\big]\,dx.

The $a$ and $b$ are usually where the graphs meet — find them by solving $f(x) = g(x)$ first.

Worked example 14.6 · Area between a curve and a line

Find the area enclosed between the curve $y = 4x - x^2$ and the $x$ -axis.

Find where the curve meets the axis ( $y = 0$ ):
$4x - x^2 = x(4 - x) = 0 \;\;\Longrightarrow\;\; x = 0 \text{ or } x = 4$
Integrate between those limits:
$\int_0^4 (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = \left(32 - \frac{64}{3}\right) - 0 = \frac{32}{3}$

Common mistake

If part of the region dips below the $x$ -axis, its integral comes out negative. Splitting the integral at the roots and taking the size of each piece avoids positive and negative areas cancelling — a classic trap in “total area” questions.

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [3 marks]

Find:

$\displaystyle\int x^3\,dx$
$\displaystyle\int 6x\,dx$
$\displaystyle\int 4\,dx$

Show solution

(a) $\dfrac{x^4}{4} + c$ .

(b) $\dfrac{6x^2}{2} + c = 3x^2 + c$ .

(c) $4x + c$ .

2 [3 marks]

Find $\displaystyle\int (4x^3 - 6x^2 + 2)\,dx$ .

Show solution

\int (4x^3 - 6x^2 + 2)\,dx = x^4 - 2x^3 + 2x + c

3 [3 marks]

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

Show solution

\big[x^3 + x\big]_0^2 = (8 + 2) - 0 = 10

4 [2 marks]

Find $\displaystyle\int \frac{5}{x^2}\,dx$ .

Show solution

Rewrite as $\displaystyle\int 5x^{-2}\,dx = \frac{5x^{-1}}{-1} + c = -\frac{5}{x} + c$ .

Standard

5 [3 marks]

Find $\displaystyle\int \left(2\sqrt{x} + \frac{1}{x^3}\right)dx$ .

Show solution

Rewrite as $\displaystyle\int \left(2x^{1/2} + x^{-3}\right)dx$ :

= \frac{2x^{3/2}}{3/2} + \frac{x^{-2}}{-2} + c = \frac{4}{3}x^{3/2} - \frac{1}{2x^2} + c

6 [4 marks]

Evaluate $\displaystyle\int_1^4 (6\sqrt{x})\,dx$ .

Show solution

$\displaystyle\int 6x^{1/2}\,dx = \frac{6x^{3/2}}{3/2} = 4x^{3/2}$ .
Apply the limits:
$\big[4x^{3/2}\big]_1^4 = 4(4^{3/2}) - 4(1) = 4(8) - 4 = 28$

7 [4 marks]

The area under $y = x^2$ between $x = 0$ and $x = k$ is $9$ . Find the value of $k$ .

Show solution

\int_0^k x^2\,dx = \left[\frac{x^3}{3}\right]_0^k = \frac{k^3}{3}

Set equal to $9$ : $\dfrac{k^3}{3} = 9 \Rightarrow k^3 = 27 \Rightarrow k = 3$ .

8 [4 marks]

A curve passes through $(2, 5)$ and has $\dfrac{dy}{dx} = 3x^2 - 4x$ . Find the equation of the curve.

Show solution

Integrate: $y = x^3 - 2x^2 + c$ .
Use $(2, 5)$ : $5 = 8 - 8 + c \Rightarrow c = 5$ .

So $y = x^3 - 2x^2 + 5$ .

9 [5 marks]

Find the area enclosed between the curve $y = 6x - x^2$ and the $x$ -axis.

Show solution

Roots: $6x - x^2 = x(6 - x) = 0 \Rightarrow x = 0$ or $x = 6$ .
Integrate between them:
$\int_0^6 (6x - x^2)\,dx = \left[3x^2 - \frac{x^3}{3}\right]_0^6 = (108 - 72) - 0 = 36$

Exam style

10 [6 marks]

The curve $C$ has equation $y = x^3 - 4x$ .

Find the coordinates of the points where $C$ meets the $x$ -axis. [2]
Evaluate $\displaystyle\int_0^2 (x^3 - 4x)\,dx$ . [2]
Explain why your answer to (b) is negative, and state the actual area between $C$ and the $x$ -axis from $x = 0$ to $x = 2$ . [2]

Show solution

(a) $x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2) = 0$ , so $x = 0, 2, -2$ — points $(0,0)$ , $(2,0)$ , $(-2,0)$ .

(b) $\left[\dfrac{x^4}{4} - 2x^2\right]_0^2 = (4 - 8) - 0 = -4$ .

(c) Between $x = 0$ and $x = 2$ the curve lies below the $x$ -axis, so the integral is negative. The actual area is its size: $4$ .

11 [6 marks]

The curve $y = x^2 + 2$ and the line $y = 3x$ intersect at two points.

Find the $x$ -coordinates of the points of intersection. [2]
Find the area enclosed between the curve and the line. [4]

Show solution

(a) Set equal: $x^2 + 2 = 3x \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow (x-1)(x-2) = 0$ , so $x = 1$ and $x = 2$ .

(b) Between these, the line $3x$ is above the curve, so integrate (line − curve):

\int_1^2 \big[3x - (x^2 + 2)\big]\,dx = \int_1^2 (-x^2 + 3x - 2)\,dx

= \left[-\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_1^2 = \left(-\tfrac{8}{3} + 6 - 4\right) - \left(-\tfrac{1}{3} + \tfrac{3}{2} - 2\right) = \tfrac{1}{6}

12 [5 marks]

Given that $\dfrac{dy}{dx} = \dfrac{12}{x^2} + 2$ and the curve passes through $(2, 7)$ , find $y$ in terms of $x$ .

Show solution

Rewrite and integrate:
$y = \int \left(12x^{-2} + 2\right)dx = \frac{12x^{-1}}{-1} + 2x + c = -\frac{12}{x} + 2x + c$
Use $(2, 7)$ : $7 = -\dfrac{12}{2} + 4 + c = -6 + 4 + c = -2 + c \Rightarrow c = 9$ .

So $y = -\dfrac{12}{x} + 2x + 9$ .