8.3 The area between two curves — Practise

1. The area

Answer: Set $x^2 = x + 2$: $x^2 - x - 2 = 0$, so $(x - 2)(x + 1) = 0$ and $x = -1$ or $x = 2$. The line is above the curve, so the area is $\displaystyle\int_{-1}^{2}(x + 2 - x^2)\,dx = \left[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\right]_{-1}^{2} = \dfrac{9}{2}$ square units.

2. The area

Answer: The intersections are $x = \pm 2$. The line is above the curve, so the area is $\displaystyle\int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2} = \dfrac{32}{3}$ square units.

3. The area

Answer: Set $x^2 = 2x$: $x(x - 2) = 0$, so $x = 0$ or $x = 2$. The line is above the curve, so the area is $\displaystyle\int_0^2 (2x - x^2)\,dx = \left[x^2 - \dfrac{x^3}{3}\right]_0^2 = \dfrac{4}{3}$ square units.

4. The area

Answer: Between $0$ and $1$ the line is above the curve, so the area is $\displaystyle\int_0^1 (x - x^3)\,dx = \left[\dfrac{x^2}{2} - \dfrac{x^4}{4}\right]_0^1 = \dfrac{1}{4}$ square units.

5. The area

Answer: The intersections are $x = 0$ and $x = 1$. The square root curve is above the line, so the area is $\displaystyle\int_0^1 \left(x^{1/2} - x\right)dx = \left[\dfrac{2}{3}x^{3/2} - \dfrac{x^2}{2}\right]_0^1 = \dfrac{1}{6}$ square units.

6. The area

Answer: The curve crosses the axis at $x = \pm 2$ and lies below it between them. The integral is $\displaystyle\int_{-2}^{2}(x^2 - 4)\,dx = -\dfrac{32}{3}$, so the area is $\dfrac{32}{3}$ square units.

7. The area

Answer: Set $6 - x^2 = 2$: $x^2 = 4$, so $x = \pm 2$. The curve is above the line, so the area is $\displaystyle\int_{-2}^{2}(4 - x^2)\,dx = \dfrac{32}{3}$ square units.

8. The area

Answer: Set $x^2 = 4x - 3$: $x^2 - 4x + 3 = 0$, so $(x - 1)(x - 3) = 0$ and $x = 1$ or $x = 3$. The line is above the curve, so the area is $\displaystyle\int_1^3 (4x - 3 - x^2)\,dx = \dfrac{4}{3}$ square units.

9. The area

Answer: Set $x^2 - 2x = x$: $x^2 - 3x = 0$, so $x(x - 3) = 0$ and $x = 0$ or $x = 3$. The line is above the curve, so the area is $\displaystyle\int_0^3 (3x - x^2)\,dx = \left[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_0^3 = \dfrac{9}{2}$ square units.

10. The total area

Answer: Factorising gives $x(x - 2)(x + 2)$, so the crossings are at $x = -2$, $0$ and $2$. The two regions have equal size. Each has area $4$, so the total area is $8$ square units.

11. The area

Answer: Set $4 - x^2 = x + 2$: $x^2 + x - 2 = 0$, so $(x + 2)(x - 1) = 0$ and $x = -2$ or $x = 1$. The curve is above the line, so the area is $\displaystyle\int_{-2}^{1}(2 - x - x^2)\,dx = \dfrac{9}{2}$ square units.

12. The area

Answer: Set $4 - x^2 = 1$: $x^2 = 3$, so $x = \pm\sqrt{3}$. The curve is above the line, so the area is $\displaystyle\int_{-\sqrt{3}}^{\sqrt{3}}(3 - x^2)\,dx = \left[3x - \dfrac{x^3}{3}\right]_{-\sqrt{3}}^{\sqrt{3}} = 4\sqrt{3}$ square units.