<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand\ds{\displaystyle}
\usepackage{epsfig}
\begin{document}

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QUESTION

\begin{description}

\item[(a)]
Using partial fractions evaluate $\int_1^2\frac{1}{x(x+2)}\,dx.$

\item[(b)]
A double integral is defined by

$$\int_{y=0}^{y=1}\!\int_{x=\sqrt{y}}^{x=1}\exp(x^3)\,dx\,dy.$$

\begin{description}

\item[(i)]
Sketch the region of integration.

\item[(ii)]
Rewrite the integral so that the integration with respect to $y$
may be performed first, and hence evaluate the integral.

\end{description}

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(a)]

$\ds\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}=\frac{A(x+2)+Bx}{x(x+2)}$

Therefore $\ds1=A(x+2)+Bx.\\ x=0;\ 1=2A+0,\ A=\frac{1}{2}\\x=-2;\
1=0-2B,\ B=-\frac{1}{2}$\\


 Therefore
 \begin{eqnarray*}
 \int_1^2\frac{1}{x(x+2)}\,dx&amp;=&amp; \int_1^2\left\{\frac{\frac{1}{2}}{x}-\frac{\frac{1}{2}}{x+2}\right\}\,dx\\
 &amp;=&amp;\left[\frac{1}{2}\ln x-\frac{1}{2}\ln (x+2)\right]_1^2\\
 &amp;=&amp;\left(\frac{1}{2}\ln 2-\frac{1}{2}\ln 4\right)-\left(\frac{1}{2}\ln
 1-\frac{1}{2}\ln 3\right)\\
 &amp;=&amp;\frac{1}{2}\ln\left(\frac{2\times3}{4}\right)\\
 &amp;=&amp;\frac{1}{2}\ln\left(\frac{3}{2}\right)
 \end{eqnarray*}

\item[(b)]

$\ds\int_{y=0}^{y=1}\!\int_{x=\sqrt{y}}^{x=1}e^{x^3}\,dx\,dy$

\begin{description}

\item[(i)]
$x=\sqrt{y},\ y=x^2,\ x=1,\ y=0,\ y=1;$

\epsfig{file=160-2000-8.eps, width=40mm}

\item[(ii)]
$$$$
 \epsfig{file=160-2000-9.eps, width=40mm}

$\ds\int_{x=0}^{x=1}\!\int_{y=0}^{y=x^2}e^{x^3}\,dy\,dx\\
I=\ds\int_{x=0}^1\left[ye^{x^3}\right]_{y=0}^{y=x^2}\,dx
=\int_{x=0}^1x^2e^{x^3}\,dx=
\left[\frac{1}{3}e^{x^3}\right]_0^1=\frac{1}{3}(e-1)$

\end{description}

\end{description}



\end{document}
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