NST First Year Maths

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 $x = \frac{1}{2} \left( 3-u \right )$ $latex \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}…

25th Jan 2022

2019 Paper 2 Question 16

(a) $I = \int\int_S (z+y^3) dS = I_1 + I_2 + I_3$ where $I_1$ is the integral over the upper circular surface, $S_1$ , at $z=b$ , $I_2$ is the integral over the cylindrical surface, $S_2$ , and $I_3$ is the integral over the lower hemi-spherical surface, $S_3$ . Each integral…