8.2 Definite integrals and the area under a curve — Practise with help

Work each question closed book first. Sketch the curve over the interval before you decide whether the integral is the area.

Warm-up: remember

Find $\displaystyle\int x^3dx$ .
Find $\displaystyle\int e^{2x}dx$ .
Evaluate $\displaystyle\int_0^1 2xdx$ .

Answers

$\dfrac{x^4}{4} + C$ .
$\dfrac{1}{2}e^{2x} + C$ .
$\Big[x^2\Big]_0^1 = 1 - 0 = 1$ .

If question 1 was hard, revisit the power rule in Section 8.1.

Practise with help

1. Evaluate $\displaystyle\int_0^3 x^2dx$ .

Hint

Find an antiderivative of $x^2$ first, then substitute the two limits.

Next step

The antiderivative is $\dfrac{x^3}{3}$ . Evaluate it at $x = 3$ and at $x = 0$ , then subtract.

Full working

\int_0^3 x^2dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9.

The curve $y = x^2$ lies above the axis on this interval, so the value $9$ is the area under the curve.

2. Evaluate $\displaystyle\int_1^2 (4x - 1)dx$ .

Hint

Integrate term by term. The antiderivative has two terms, and each is evaluated at both limits.

Next step

An antiderivative is $2x^2 - x$ . Evaluate at $x = 2$ and at $x = 1$ .

Full working

\int_1^2 (4x - 1)dx = \Big[2x^2 - x\Big]_1^2 = (8 - 2) - (2 - 1) = 6 - 1 = 5.

Take care with the subtraction: the second bracket is $1$ , so the answer is $5$ .

3. Evaluate $\displaystyle\int_0^1 \left(e^x + 1\right)dx$ .

Hint

The antiderivative of $e^x$ is $e^x$ . Keep $e$ as an exact value.

Next step

An antiderivative is $e^x + x$ . Evaluate at $1$ and at $0$ , remembering that $e^0 = 1$ .

Full working

\int_0^1 \left(e^x + 1\right)dx = \Big[e^x + x\Big]_0^1 = (e + 1) - (1 + 0) = e.

The two $1$ s cancel, so the exact answer is $e \approx 2.718$ .

4. Find the area under the curve $y = 9 - x^2$ between $x = 0$ and $x = 3$ .

Hint

Check that the curve is above the axis on the interval. At the lower end $y = 9$ and at the upper end $y = 0$ , and the curve falls steadily.

Next step

The area is $\displaystyle\int_0^3 (9 - x^2)dx$ . An antiderivative is $9x - \dfrac{x^3}{3}$ .

Full working

A = \int_0^3 (9 - x^2)dx = \left[9x - \frac{x^3}{3}\right]_0^3 = (27 - 9) - 0 = 18.

The area is $18$ square units. The curve touches the axis at $x = 3$ but stays above it, so the whole integral is the area.

5. Evaluate $\displaystyle\int_{-1}^1 x^3dx$ and explain your answer.

Hint

Find the antiderivative, then substitute both limits. Look at the symmetry of the curve $y = x^3$ .

Next step

The antiderivative is $\dfrac{x^4}{4}$ . Evaluate at $1$ and at $-1$ .

Full working

\int_{-1}^1 x^3dx = \left[\frac{x^4}{4}\right]_{-1}^1 = \frac{1}{4} - \frac{1}{4} = 0.

The curve is odd: the region below the axis between $-1$ and $0$ has the same size as the region above the axis between $0$ and $1$ . The signed areas cancel, so the integral is $0$ . The total area is $2 \times \dfrac{1}{4} = \dfrac{1}{2}$ square units.

6. Find the total area between the curve $y = x^2 - 1$ and the horizontal axis, from $x = 0$ to $x = 2$ .

Hint

Find where the curve crosses the axis. Solve $x^2 - 1 = 0$ for $x > 0$ .

Next step

The curve crosses at $x = 1$ , so split the range into $0$ to $1$ and $1$ to $2$ . Take the size of the first integral and add the second.

Full working

An antiderivative is $\dfrac{x^3}{3} - x$ .

\int_0^1 (x^2 - 1)dx = \left(\frac{1}{3} - 1\right) - 0 = -\frac{2}{3}, \qquad \int_1^2 (x^2 - 1)dx = \left(\frac{8}{3} - 2\right) - \left(\frac{1}{3} - 1\right) = \frac{4}{3}.

The first part lies below the axis, so take its size. The total area is

\frac{2}{3} + \frac{4}{3} = 2

square units.