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{\bf Exam Question

Topic: Double Integral}

Let $I$ denote the repeated integral $$\int_0^1\, dx
\int_0^{2x-x^2}\exp\left[(1-y)^{3/2}\right]\, dy.$$

Reverse the order of integration and hence evaluate $I$. Give your
answer in terms of e, and also as a decimal correct to 3 places.

\vspace{0.5in}

{\bf Solution}

Now $y=2x-x^2\Rightarrow x=1\pm\sqrt{1-y}.$ But $0\le x\le1$ and
so $x=1-\sqrt{1-y}.$

So reversing the order of integration gives
\begin{eqnarray*}
I&=&\int_0^1\,
dy\int_{1-\sqrt{1-y}}^1\exp\left[(1-y)^{3/2}\right]\, dx
=\int_0^1\left[x\exp(1-y)^{3/2}\right]_{1-\sqrt{1-y}}^1\\ &=&
\int_0^1\sqrt{1-y}\, \exp\left[(1-y)^{3/2}\right]\, dy =
\left[\textstyle{\frac{2}{3}}(-1)\exp(1-y)^{3/2}\right]_0^1\\
&=&\textstyle{\frac{2}{3}}(-1+\mathrm{e})=1.146\ \ \mathrm{3\
d.p.}
\end{eqnarray*}
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