2024 H2 Mathematics Paper 1 Question 10

Definite Integrals: Areas and Volumes

Answers

$\int \frac{2 u^{2}}{u^{2} - 9} d u .$

(b)

Sketch.

$1 + 3 \ln \frac{8}{5} .$

$11.04 {units}^{2}$

Full Solutions

(a)

\begin{aligned} u & = \sqrt{4 x + 1} \\ \frac{d u}{d x} & = \frac{2}{\sqrt{4 x + 1}} \\ = \frac{2}{u} \end{aligned}

\begin{aligned} u & = \sqrt{4 x + 1} \\ u^{2} & = 4 x + 1 \\ 4 x & = u^{2} - 1 \\ x & = \frac{u^{2} - 1}{4} \end{aligned}

\begin{aligned} \int \frac{\sqrt{4 x + 1}}{x - 2} d x \\ = \int \frac{u}{\frac{u^{2} - 1}{4} - 2} \frac{u}{2} d u \\ = \int \frac{u^{2}}{\frac{u^{2} - 1}{2} - 4} d u \\ = \int \frac{2 u^{2}}{u^{2} - 9} d u ∎ \end{aligned}

(b)

(i)

graph

(ii)

When $x = 6,$ $u = 5$
When $x = 12,$ $u = 7$

\begin{aligned} Area under curve \\ = \int_{6}^{12} \frac{\sqrt{4 x + 1}}{x - 2} d x \\ = \int_{5}^{7} \frac{2 u^{2}}{u^{2} - 9} d u \\ = \int_{5}^{7} (2 + \frac{18}{u^{2} - 9}) d u \\ = {[2 u + 3 \ln | \frac{u - 3}{u + 3} |]}_{5}^{7} \\ = 2 (7) + 3 \ln | \frac{7 - 3}{7 + 3} | - (2 (5) + 3 \ln | \frac{5 - 3}{5 + 3} |) \\ = 4 + 3 \ln \frac{2}{5} - 3 \ln \frac{1}{4} \\ = 4 + 3 \ln \frac{8}{5} \end{aligned}

\begin{aligned} Area of R \\ = Area under curve - area of rectangle \\ = 4 + 3 \ln \frac{8}{5} - ((12 - 6) (0.5)) \\ = 1 + 3 \ln \frac{8}{5} ∎ \end{aligned}

(iii)

\begin{aligned} Volume required \\ π \int_{6}^{12} {(\frac{\sqrt{4 x + 1}}{x - 2})}^{2} d x - volume of cylinder \\ = π \int_{6}^{12} {(\frac{\sqrt{4 x + 1}}{x - 2})}^{2} d x - (π (0 . 5^{2}) (12 - 6)) \\ = 11.04 {units}^{2} (2 d.p.) ∎ \end{aligned}