2007 H2 Mathematics Paper 2 Question 4

Definite Integrals: Areas and Volumes

Answers

(i)

∫05π3sin⁡2x  dx=56π+183{\displaystyle \int_0^{\frac{5 \pi}{3}} \sin^2 x \; \mathrm{d}x} \allowbreak {\displaystyle = \frac{5}{6} \pi + \frac{1}{8} \sqrt{3}}
∫05π3cos⁡2x  dx=56π−183{\displaystyle \int_0^{\frac{5 \pi}{3}} \cos^2 x \; \mathrm{d}x} \allowbreak {\displaystyle = \frac{5}{6} \pi - \frac{1}{8} \sqrt{3}}
(iia)
(π−2) units2{(\pi - 2) \textrm{ units}^2}
(iib)
5.391 units3{5.391 \textrm{ units}^3}

Full solutions

(i)

=∫05π3sin⁡2x  dx=∫05π31−cos⁡2x2  dx=[x2−sin⁡2x4]05π3=12(5π3)−14(sin⁡10π3−sin⁡0)=56π+183  ■\begin{align*} & = \int_0^\frac{5 \pi}{3} \sin^2 x \; \mathrm{d}x \\ & = \int_0^\frac{5 \pi}{3} \frac{1-\cos 2x}{2} \; \mathrm{d}x \\ & = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_0^{\frac{5 \pi}{3}}\\ & = \frac{1}{2} \left( \frac{5 \pi}{3} \right) - \frac{1}{4} \left( \sin \frac{10\pi}{3} - \sin 0 \right) \\ & = \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \; \blacksquare \end{align*}
=∫05π3cos⁡2x  dx=∫05π31−sin⁡2x  dx=[x]05π3−(56π+183)=5π3−(56π+183)=56π−183  ■\begin{align*} & = \int_0^\frac{5 \pi}{3} \cos^2 x \; \mathrm{d}x \\ & = \int_0^\frac{5 \pi}{3} 1 - \sin^2 x \; \mathrm{d}x \\ & = \Big[ x \Big]_0^\frac{5 \pi}{3} - \left( \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \right) \\ & = \frac{5 \pi}{3} - \left( \frac{5}{6} \pi + \frac{1}{8} \sqrt{3} \right) \\ & = \frac{5}{6} \pi - \frac{1}{8} \sqrt{3} \; \blacksquare \end{align*}
(iia)
∫012πx2sin⁡x  dx=[x2(−cos⁡x)]012π−∫012π2x(−cos⁡x)  dx=0−0+∫012π2xcos⁡x  dx=[2xsin⁡x]012π−∫012π2sin⁡x  dx=2(12π)−0+[2cos⁡x]012π=π+2cos⁡12π−2cos⁡0=(π−2) units2  ■\begin{align*} & \int_0^{\frac{1}{2}\pi} x^{2} \sin x \; \mathrm{d}x \\ & = \Big[ x^2 (- \cos x) \Big]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} 2 x (- \cos x) \; \mathrm{d}x \\ & = 0 - 0 + \int_0^{\frac{1}{2}\pi} 2 x \cos x \; \mathrm{d}x \\ & = \Big[ 2x \sin x \Big]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} 2 \sin x \; \mathrm{d}x \\ & = 2 \left( \frac{1}{2}\pi \right) - 0 + \Big[ 2 \cos x \Big]_0^{\frac{1}{2}\pi} \\ & = \pi + 2 \cos \frac{1}{2}\pi - 2\cos 0 \\ & = (\pi - 2) \textrm{ units}^2 \; \blacksquare \\ \end{align*}
(iib)
Volume of revolution=π∫012π(x2sin⁡x)2  dx=5.391 units3 (3 dp)  ■\begin{align*} & \textrm{Volume of revolution} \\ & = \pi \int_0^{\frac{1}{2}\pi} \left( x^2 \sin x \right)^2 \; \mathrm{d}x \\ & = 5.391 \textrm{ units}^3 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}