2020 Paper 1 Question 6

$I = \int x \sqrt{3-2\,x}\, dx$

You can do this by substitution.

Let $u = 3 -2x$

$\Rightarrow dx = -\frac{1}{2} du$ and $x = \frac{1}{2} \left( 3-u \right )$

$\begin{aligned} \therefore I &= \int \frac{(3-u)}{2} \, u^{1/2} \, \left(-\frac{1}{2} \right) \,du \\ &= -\frac{1}{4} \int \left( 3u^{1/2} - u^{3/2} \right) \, du \\ &= -\frac{1}{4} \left( 2u^{3/2}-\frac{2}{5} u^{5/2} \right) + C \\ &= -\frac{1}{2} u^{3/2} \left( \frac{5-u}{5} \right) + C \\ &= -\frac{1}{2} (3-2x)^{3/2} \left( \frac{2+2x}{5} \right) + C \\ &= -\frac{1}{5} (3-2x)^{3/2} \left( 1+x \right) + C \end{aligned}$

This is probably the most compact form. Other forms are possible, for example by writing $(3-2x)^{3/2} = (3-2x)^{1/2}(3-2x)$ .

NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

2020 Paper 1 Question 6

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