2021 H2 Mathematics Paper 1 Question 9

Integration Techniques

Answers

(a)

Maximum point (π4,22eπ4){\left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \mathrm{e}^{\frac{\pi}{4}} \right)}

(b)

14e2x(sin⁡2x+cos⁡2x)+c{\frac{1}{4} \mathrm{e}^{2 x} (\sin 2 x+\cos 2 x) + c}

(c)

π8(eπ−3){\frac{\pi}{8} \left( \mathrm{e}^{\pi} - 3 \right)}

Full solutions

(a)

f′(x)=excos⁡x−exsin⁡x=ex(cos⁡x−sin⁡x)\begin{align*} f'(x) &= \mathrm{e}^{x} \cos x - \mathrm{e}^{x} \sin x \\ &= \mathrm{e}^{x} (\cos x - \sin x) \\ \end{align*}
At stationary point, f′(x)=0{f'(x)=0}
ex(cos⁡x−sin⁡x)=0cos⁡x−sin⁡x=0sin⁡x=cos⁡xtan⁡x=1x=π4\begin{gather*} \mathrm{e}^{x} (\cos x - \sin x) = 0 \\ \cos x - \sin x = 0 \\ \sin x = \cos x \\ \tan x = 1 \\ x = \frac{\pi}{4} \end{gather*}
f′(x)=f(x)−exsin⁡xf′′(x)=f′(x)−exsin⁡x−excos⁡x\begin{align*} f'(x) &= f(x) - \mathrm{e}^{x} \sin x \\ f''(x) &= f'(x) - \mathrm{e}^{x}\sin x - \mathrm{e}^{x}\cos x \end{align*}
f(π4)=eπ4cos⁡π4=22eπ4f′′(π4)=0−eπ422−eπ422=−2eπ4<0\begin{align*} f\left( \frac{\pi}{4} \right) &= \mathrm{e}^{\frac{\pi}{4}} \cos \frac{\pi}{4} \\ & = \frac{\sqrt{2}}{2} \mathrm{e}^{\frac{\pi}{4}} \\ f''\left( \frac{\pi}{4} \right) &= 0 - \mathrm{e}^{\frac{\pi}{4}} \frac{\sqrt{2}}{2} - \mathrm{e}^{\frac{\pi}{4}} \frac{\sqrt{2}}{2} \\ &= - \sqrt{2} \mathrm{e}^{\frac{\pi}{4}} \\ &< 0 \end{align*}
Hence the stationary point is
(π4,22eπ4)  ■\left( \frac{\pi}{4}, \frac{\sqrt{2}}{2} \mathrm{e}^{\frac{\pi}{4}} \right) \; \blacksquare
and it is a maximum point ■{\blacksquare}

(b)

∫e2xcos⁡2x  dx=e2x2cos⁡2x−∫e2x2(−2sin⁡2x)  dx=e2xcos⁡2x2+∫e2xsin⁡2x  dx=e2xcos⁡2x2+e2x2sin⁡2x=+−∫e2x2(2cos⁡2x)  dx∫e2xcos⁡2x  dx=e2xcos⁡2x+e2xsin⁡2x2−∫e2xcos⁡2x  dx2∫e2xcos⁡2x  dx=e2xcos⁡2x+e2xsin⁡2x2+C∫e2xcos⁡2x  dx=14e2x(sin⁡2x+cos⁡2x)+c  ■\begin{align*} \int \mathrm{e}^{2 x} \cos 2 x \; \mathrm{d}x &= \frac{\mathrm{e}^{2 x}}{2} \cos 2 x - \int \frac{\mathrm{e}^{2 x}}{2} (-2\sin 2 x) \; \mathrm{d}x \\ & = \frac{\mathrm{e}^{2 x}\cos 2 x}{2} + \int \mathrm{e}^{2 x} \sin 2 x \; \mathrm{d}x \\ & = \frac{\mathrm{e}^{2 x}\cos 2 x}{2} + \frac{\mathrm{e}^{2 x}}{2} \sin 2x \\ & \phantom{= +} - \int \frac{\mathrm{e}^{2 x}}{2} (2 \cos 2 x) \; \mathrm{d}x \\ \int \mathrm{e}^{2 x} \cos 2 x \; \mathrm{d}x & = \frac{\mathrm{e}^{2 x}\cos 2 x+\mathrm{e}^{2 x}\sin 2 x}{2} - \int \mathrm{e}^{2 x} \cos 2 x \; \mathrm{d}x \\ 2 \int \mathrm{e}^{2 x} \cos 2 x \; \mathrm{d}x & = \frac{\mathrm{e}^{2 x}\cos 2 x+\mathrm{e}^{2 x}\sin 2 x}{2} + C \\ \int \mathrm{e}^{2 x} \cos 2 x \; \mathrm{d}x & = \frac{1}{4} \mathrm{e}^{2 x} (\sin 2 x+\cos 2 x) + c \; \blacksquare \end{align*}

(c)

At x-{x\textrm{-}}intercepts,
excos⁡x=0cos⁡x=0x=π2\begin{gather*} \mathrm{e}^{x} \cos x = 0 \\ \cos x = 0 \\ x = \frac{\pi}{2} \end{gather*}
Volume generated=π∫0π2(excos⁡x)2  dx=π∫0π2e2xcos⁡2x  dx=π∫0π2e2xcos⁡2x+12  dx=π2∫0π2e2xcos⁡2x+e2x  dx=π2[14e2x(sin⁡2x+cos⁡2x)+e2x2]0π2=π2(14eπ(0−1)+12eπ−14(0+1)−12)=π2(14eπ−34)=π8(eπ−3) units3  ■\begin{align*} & \textrm{Volume generated} \\ & = \pi \int_0^{\frac{\pi}{2}} \left( \mathrm{e}^{x} \cos x \right)^2 \; \mathrm{d}x \\ & = \pi \int_0^{\frac{\pi}{2}} \mathrm{e}^{2 x} \cos^2 x \; \mathrm{d}x \\ & = \pi \int_0^{\frac{\pi}{2}} \mathrm{e}^{2 x} \frac{\cos 2x + 1}{2} \; \mathrm{d}x \\ & = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \mathrm{e}^{2 x} \cos 2x + \mathrm{e}^{2 x} \; \mathrm{d}x \\ & = \frac{\pi}{2} \left[ \frac{1}{4} \mathrm{e}^{2 x} (\sin 2 x+\cos 2 x) + \frac{\mathrm{e}^{2 x}}{2} \right]_0^{\frac{\pi}{2}} \\ & = \frac{\pi}{2} \left( \frac{1}{4} \mathrm{e}^{\pi} (0-1) + \frac{1}{2} \mathrm{e}^{\pi} - \frac{1}{4}(0+1) - \frac{1}{2} \right) \\ & = \frac{\pi}{2} \left( \frac{1}{4} \mathrm{e}^{\pi} - \frac{3}{4} \right) \\ & = \frac{\pi}{8} \left( \mathrm{e}^{\pi} - 3 \right) \textrm{ units}^3 \; \blacksquare \end{align*}