2018 H2 Mathematics Paper 1 Question 2

Definite Integrals: Areas and Volumes

Answers

{x=\frac{1}{2}}

and

{x=3}

{ \frac{125\pi}{6} \textrm{ units}^3 }

\begin{gather*} \frac{3}{x} = 7 - 2x \\ 3 = 7x - 2x^2 \\ 2 x^2 - 7 x + 3 = 0 \\ (2 x - 1)(x - 3) = 0 \end{gather*}

{x}

-coordinates of

{A}

and

{B}

x = \frac{1}{2}, \quad x = 3 \; \blacksquare

\begin{align*} & \textrm{Exact volume generated} \\ & = \pi \int_{\frac{1}{2}}^{3} \left( (7-2x)^2 - \left( \frac{3}{x} \right)^2 \right) \; \mathrm{d}x \\ & = \pi \int_{\frac{1}{2}}^{3} \left( (7 - 2 x)^2 - \frac{9}{x^{2}} \right) \; \mathrm{d}x \\ & = \pi \left[ - \frac{1}{6} \left(7 - 2 x\right)^{3} + \frac{9}{x} \right]_{\frac{1}{2}}^{3} \\ &= \pi \left( -\frac{1}{6} (7-6)^3 + \frac{9}{3} + \frac{1}{6} (7-1)^3 - 18 \right) \\ &= \pi \left( -\frac{1}{6} - 15 + 36 \right) \\ &= \frac{125\pi}{6} \textrm{ units}^3 \; \blacksquare \\ \end{align*}