Test

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Question 12 A particle $P$ travels in a straight line so that, $t$ seconds after passing through a fixed point $O$, its velocity, $v \mathrm{~ms}^{-1}$, is given by $v=\dfrac{t}{2 \mathrm{e}}$ for $0 \leqslant t \leqslant 2$, $v=\mathrm{e}^{-\dfrac{t}{2}} \quad$ for $t>2$. Given that, after leaving $O$, particle $P$ is never at rest, find the distance it travels between $t=1$ and $t=3 .$
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*********math solution************* $\begin{array}{ll} &\text { The distance it travels between } t=1 \text { and } t=3 \\&=\int_{1}^{2} \dfrac{t}{2 e} d t+\int_{2}^{3} e^{-\frac{t}{2}} d t \\&=\frac{1}{2e} \left[\dfrac{t^{2}}{2} \right]_{1}^{2}+\left[ \dfrac{e^{- \dfrac{t}{2}}}{\left(-\dfrac{1}{2}\right)}\right]_{2}^{3} \\&=\dfrac{1}{4 e}\left(2^{2}-1^{2}\right)-2\left(e^{-\dfrac{3}{2}}-e^{-1}\right) \\&=0.565\end{array}$ **********end math solution********************

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Question $\begin{array}{lll}\mathbf{11} & \text { (a) (i) Find } & \int\dfrac{1}{(10 x-1)^{6}} \mathrm{~d} x \text { . }\end{array}$ (ii) Find $\int \dfrac{\left(2 x^{3}+5\right)^{2}}{x} \mathrm{~d} x$. (b) (i) Differentiate $y=\tan (3 x+1)$ with respect to $x$. (ii) Hence find $\int_{\dfrac{\pi}{12}}^{\dfrac{\pi}{10}}\left(\dfrac{\sec ^{2}(3 x+1)}{2}-\sin x\right) \mathrm{d} x$.
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*********math solution************* $\begin{array}{ll}(a)(i) \displaystyle\int \frac{1}{(10 x-1)^{6}} d x &=\int(10 x-1)^{-6} d x \\&=\frac{1}{10} \times \frac{(10 x-1)^{-5}}{-5}+c \\&=-\frac{1}{50}(10 x-1)^{-5}+c \\\text { (ii) } \displaystyle\int \frac{\left(2 x^{3}+5\right)^{2}}{x} d x &=\int \frac{4 x^{6}+20 x^{3}+25}{x} d x \\&=\int\left(4 x^{5}+20 x^{2}+\frac{25}{x}\right) d x \\&=4 \frac{x^{6}}{6}+20 \frac{x^{3}}{3}+25 \ln x+c\end{array}$ b(i) $\dfrac{dy}{dx}=3\sec^2(3x+1)$ \text { (b) (ii) } $\begin{array}[t]{ll}\displaystyle\int_{\frac{\pi}{12}}^{\frac{\pi}{10}} &\left(\dfrac{\sec ^{2}(3 x+1)}{2}-\sin x\right) d x \\=& \dfrac{1}{2} \left[\dfrac{\tan (3 x+1)}{3}\right]_{\frac{\pi}{12}}^{\frac{\pi}{10}}+\left[\cos x\right]_{\frac{\pi}{12}}^{\frac{\pi}{10}} \\=& \dfrac{1}{6}\left[\tan \left(\dfrac{3 \pi}{10}+1\right)-\tan \left(\dfrac{3 \pi}{12}+1\right)\right] \\&+\left[\cos \left(\frac{\pi}{10}\right)-\cos \left(\frac{\pi}{12}\right)\right] \\=& 0.322\end{array}$ **********end math solution********************