8.3 The area between two curves — Practise with help

Work each question closed book first. Find the intersections, decide which curve is upper, then integrate upper minus lower.

Warm-up: remember

Solve $x^2 = 4$ for $x$ .
Solve $x^2 = 3x$ for $x$ .
Evaluate $\displaystyle\int_0^1 \left(x - x^2\right)dx$ .

Answers

$x = 2$ or $x = -2$ .
$x = 0$ or $x = 3$ .
$\left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1 = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$ .

If question 2 was hard, revise solving quadratics by factorising in Section 2.4.

Practise with help

1. Find the area enclosed between the curve $y = x^2$ and the line $y = 4$ .

Hint

Find the intersections by solving $x^2 = 4$ . Sketch both graphs to see which is the upper curve.

Next step

The intersections are at $x = -2$ and $x = 2$ . The line is above the curve, so integrate $(4 - x^2)$ between the limits.

Full working

A = \int_{-2}^{2} (4 - x^2)dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}.

The area is $\dfrac{32}{3}$ square units.

2. Find the area enclosed between the curve $y = x^2$ and the line $y = 3x$ .

Hint

Set $x^2 = 3x$ and factorise. One intersection is at the origin.

Next step

$x(x - 3) = 0$ gives $x = 0$ and $x = 3$ . The line is above the curve between them, so integrate $(3x - x^2)$ .

Full working

A = \int_0^3 (3x - x^2)dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}.

The area is $\dfrac{9}{2}$ square units.

3. Find the area enclosed between the curve $y = x^2 - 2$ and the line $y = 2$ .

Hint

Solve $x^2 - 2 = 2$ . The line is the upper curve, because the parabola opens upwards and its turning point is below the line.

Next step

The intersections are at $x = -2$ and $x = 2$ . The height is $2 - (x^2 - 2) = 4 - x^2$ .

Full working

A = \int_{-2}^{2} (4 - x^2)dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{32}{3}.

The translation of the curve does not change the shape of the enclosed region, so the area matches question 1.

4. Find the area between the curve $y = x^3$ and the line $y = x$ , from $x = 0$ to $x = 1$ .

Hint

Over this interval the two graphs meet only at the ends. Test a point in the middle, such as $x = 0.5$ , to decide which is upper.

Next step

At $x = 0.5$ the line gives $0.5$ and the curve gives $0.125$ , so the line is upper. Integrate $(x - x^3)$ from $0$ to $1$ .

Full working

A = \int_0^1 (x - x^3)dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.

The area is $\dfrac{1}{4}$ square units.

5. Find the area enclosed between the curve $y = \sqrt{x}$ and the line $y = x$ .

Hint

Solve $\sqrt{x} = x$ . Square both sides, then check that the solutions you find are genuine.

Next step

Squaring gives $x = x^2$ , so $x = 0$ or $x = 1$ . Both satisfy the original equation. The square root curve is above the line between them.

Full working

A = \int_0^1 \left(x^{1/2} - x\right)dx = \left[\frac{2}{3}x^{3/2} - \frac{x^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}.

The area is $\dfrac{1}{6}$ square units.

6. Find the total area enclosed between the curve $y = x^3 - 4x$ and the horizontal axis.

Hint

Factorise $x^3 - 4x = x(x - 2)(x + 2)$ to find the three crossings. The region has symmetry.

Next step

The crossings are at $x = -2$ , $0$ and $2$ . The area from $0$ to $2$ equals the area from $-2$ to $0$ , so find one and double it.

Full working

\int_0^2 (x^3 - 4x)dx = \left[\frac{x^4}{4} - 2x^2\right]_0^2 = 4 - 8 = -4.

The curve lies below the axis between $0$ and $2$ , so the area of that part is $4$ . By symmetry, the total area is

4 + 4 = 8

square units.