8.2 Definite integrals and the area under a curve — Practise

Work through these on your own. For each area question, sketch the curve first and decide whether the integral is positive, negative or split.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Evaluate $\displaystyle\int_0^2 x^3\,dx$ .
Evaluate $\displaystyle\int_1^3 (2x + 1)\,dx$ .
Evaluate $\displaystyle\int_0^1 (3x^2 - 2x + 1)\,dx$ .
Evaluate $\displaystyle\int_1^2 \frac{1}{x}\,dx$ , giving an exact answer.
Evaluate $\displaystyle\int_0^1 e^{2x}\,dx$ , giving an exact answer.
Evaluate $\displaystyle\int_0^{\pi/2} \cos x\,dx$ .
Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$ .
Evaluate $\displaystyle\int_{-1}^{1} x^3\,dx$ and explain the value.
Evaluate $\displaystyle\int_0^3 (x^2 - 4)\,dx$ .
Find the area under the curve $y = 4 - x^2$ between $x = -2$ and $x = 2$ .
Find the area under the curve $y = x^2$ between $x = 0$ and $x = 3$ .
Find the total area between the curve $y = x^2 - 2x$ and the horizontal axis, from $x = 0$ to $x = 3$ .

Answers and concise solutions

1. The value of the integral

Answer: $\displaystyle\int_0^2 x^3\,dx = \left[\dfrac{x^4}{4}\right]_0^2 = 4$ .

2. The value of the integral

Answer: $\displaystyle\int_1^3 (2x + 1)\,dx = \Big[x^2 + x\Big]_1^3 = 12 - 2 = 10$ .

3. The value of the integral

Answer: $\displaystyle\int_0^1 (3x^2 - 2x + 1)\,dx = \Big[x^3 - x^2 + x\Big]_0^1 = 1 - 0 = 1$ .

4. The value of the integral

Answer: $\displaystyle\int_1^2 \dfrac{1}{x}\,dx = \Big[\ln x\Big]_1^2 = \ln 2 - \ln 1 = \ln 2$ .

5. The value of the integral

Answer: $\displaystyle\int_0^1 e^{2x}\,dx = \left[\dfrac{1}{2}e^{2x}\right]_0^1 = \dfrac{e^2 - 1}{2}$ .

6. The value of the integral

Answer: $\displaystyle\int_0^{\pi/2} \cos x\,dx = \Big[\sin x\Big]_0^{\pi/2} = 1 - 0 = 1$ .

7. The value of the integral

Answer: $\displaystyle\int_0^{\pi} \sin x\,dx = \Big[-\cos x\Big]_0^{\pi} = 1 - (-1) = 2$ .

8. The value of the integral

Answer: $\displaystyle\int_{-1}^{1} x^3\,dx = \left[\dfrac{x^4}{4}\right]_{-1}^{1} = \dfrac{1}{4} - \dfrac{1}{4} = 0$ . The region below the axis cancels the region above it, because the curve has rotational symmetry about the origin.

9. The value of the integral

Answer: $\displaystyle\int_0^3 (x^2 - 4)\,dx = \left[\dfrac{x^3}{3} - 4x\right]_0^3 = 9 - 12 = -3$ . The value is negative because the curve lies below the axis for part of the interval.

10. The area

Answer: The curve is above the axis on the whole interval, so the area is $\displaystyle\int_{-2}^2 (4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_{-2}^2 = \dfrac{16}{3} + \dfrac{16}{3} = \dfrac{32}{3}$ square units.

11. The area

Answer: $\displaystyle\int_0^3 x^2\,dx = \left[\dfrac{x^3}{3}\right]_0^3 = 9$ square units.

12. The total area

Answer: The curve crosses the axis at $x = 0$ and $x = 2$ . The first piece is $\displaystyle\int_0^2 (x^2 - 2x)\,dx = -\dfrac{4}{3}$ , and the second is $\displaystyle\int_2^3 (x^2 - 2x)\,dx = \dfrac{4}{3}$ . The total area is $\dfrac{4}{3} + \dfrac{4}{3} = \dfrac{8}{3}$ square units.