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{\bf Exam Question

Topic: Double Integral}

The region $R$ is a trapezium bounded by the lines
$x=\frac{1}{2},x=1.y=x,y=2x.$

Evaluate the double integral $$\int \!\!\! \int_R \frac{\sin
x}{x}\, d(x,y).$$

Give your answer in exact form and also as an approximation
rounded to four decimal places using your calculator.

\vspace{0.5in}

{\bf Solution}

To evaluate the double integral we integrate with respect to $y$
first, giving:

$$\int_{x=1/2}^1\, dx \int_{y=x}^{2x} \frac{\sin x}{y}\, dy =
\int_{x=1/2}^1\left[\frac{y\sin x}{x}\right]_{y=x}^{2x}\, dx$$ $$
= \int_{x=1/2}^1\sin x\, dx =\cos(1/2)-\cos(1)=0.3373\ \
{\mathrm{(4\ d.p.)}}$$




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