Solutions — Area Between Two Curves (Ex 11I)

Year 12 Mathematical Methods Unit 3 · Cambridge §11I · Mr Wong

ANSWER KEY

Warm-up — intersections

Q1. Solve each $f(x)=g(x)$:

a $x=-1, 3$ $x^2-2x-3=(x-3)(x+1)$

b $x=0, 2$ $x(x-2)=0$

c $x=-2, 1$ $x^2+x-2=(x+2)(x-1)$

d $x=\pm 2$ $2x^2=8$

Part A — Straight area between curves

2 Line on top. $\int_{-1}^{3}[(x+1)-(x^2-x-2)]\,dx = \int_{-1}^{3}(-x^2+2x+3)\,dx = [-\tfrac{x^3}{3}+x^2+3x]_{-1}^{3} = 9-(-\tfrac{5}{3}) =$ $\dfrac{32}{3}$ sq units

3 Parabola on top. $\int_{-2}^{1}[(4-x^2)-(x+2)]\,dx = \int_{-2}^{1}(2-x-x^2)\,dx = [2x-\tfrac{x^2}{2}-\tfrac{x^3}{3}]_{-2}^{1} = \tfrac{7}{6} - (-\tfrac{10}{3}) =$ $\dfrac{9}{2}$ sq units

4 Intersections $x^2=x+2 \Rightarrow x=-1,2$. Line on top. $\int_{-1}^{2}[(x+2)-x^2]\,dx = [\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}]_{-1}^{2} = \tfrac{10}{3}-(-\tfrac{7}{6}) =$ $\dfrac{9}{2}$ sq units

Part B — Curves that swap

5 Roots of $x^3-x=0$ at $-1,0,1$. On $[-1,0]$ cubic is above; on $[0,1]$ line is above. $\int_{-1}^{0}(x^3-x)\,dx = \tfrac{1}{4}$ and $\int_{0}^{1}(x-x^3)\,dx = \tfrac{1}{4}$. Total area $= \dfrac{1}{2}$ sq units

6 On $[\tfrac{\pi}{4},\tfrac{5\pi}{4}]$, $\sin x \ge \cos x$. $\int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx = [-\cos x - \sin x]_{\pi/4}^{5\pi/4} = (\tfrac{\sqrt 2}{2}+\tfrac{\sqrt 2}{2}) - (-\tfrac{\sqrt 2}{2}-\tfrac{\sqrt 2}{2}) =$ $2\sqrt{2}$ sq units

Part C — Application & VCAA-style

7 On $[0,1]$, $e^x \ge x$. $\int_0^1(e^x - x)\,dx = [e^x - \tfrac{x^2}{2}]_0^1 = (e-\tfrac{1}{2}) - 1 =$ $e - \dfrac{3}{2}$ sq units ($\approx 1.218$).

8 Intersections $x^2 = 8-x^2 \Rightarrow x = \pm 2$. The inverted parabola $y=8-x^2$ is on top. $\int_{-2}^{2}[(8-x^2)-x^2]\,dx = \int_{-2}^{2}(8-2x^2)\,dx = [8x-\tfrac{2x^3}{3}]_{-2}^{2} = (16-\tfrac{16}{3}) - (-16+\tfrac{16}{3}) =$ $\dfrac{64}{3}$ sq units

9 Intersections: $4-x^2 = mx \Rightarrow x = \tfrac{-m\pm\sqrt{m^2+16}}{2}$. With $a, b$ the smaller and larger root, $b-a = \sqrt{m^2+16}$. The integral simplifies to $\displaystyle\int_a^b (4 - mx - x^2)\,dx = \dfrac{(m^2+16)^{3/2}}{6}$. Set $=\tfrac{125}{6}$: $(m^2+16)^{3/2} = 125 \Rightarrow m^2+16 = 25 \Rightarrow m^2 = 9$. With $m>0$, $m = 3$.