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QUESTION


\begin{description}

\item[(i)]
Evaluate the double integral
$\displaystyle{\int_0^2\!\int_{y^2}^{2y}(4x-y)\,dxdy}$.

\item[(ii)]
Given the integral $\displaystyle{\int_0^4\!\int_{\sqrt{y}}^2y\cos
x^5\,dxdy}$, sketch the region of integration. Reverse the order
of integration and evaluate the resulting integral.

\item[(iii)]
Sketch the plane region $R$ defined by the inequalities:

$$y\leq\sqrt{x},\ y\geq\sqrt{3x-18},\ x\geq0.$$

By choosing an appropriate order of integration show how to
express an integral of the form $\int\!\!\!\int_Rf(x,y)\,dA$ as a
double integral, computing the limits.

\end{description}



ANSWER


\begin{description}

\item[(i)]

\begin{eqnarray*}
\int_0^2\!\int_{y^2}^{2y}(4x-y)\,dxdy&=&
\int_0^2\left[\frac{4x^2}{2}-yx\right]_{y^2}^{2y}\,dy\\
&=&\int_0^28y^2-2y^2-2y^4+y^3\,dy\\
&=&\left[\frac{2y^5}{5}+\frac{y^4}{4}+\frac{6y^3}{3}\right]_0^2=7\frac{1}{5}
\end{eqnarray*}

\item[(ii)]

DIAGRAM

\begin{eqnarray*}
\int_{x=0}^2\int_{y=0}^{x^2}y\cos
x^5\,dydx&=&\int_0^2\left[\frac{y^2}{2}\right]_0^{x^2}\cos
x^5\,dx\\ &=&\int_0^2\frac{x^4}{2}\cos x^5\,dx
\end{eqnarray*}

Putting $u=x^5$, so $\frac{du}{dx}=5x^4$ we get

$$\int_{x=0}^2\frac{\cos u}{10}\,du=\int_{u=0}^32\frac{\cos
u}{10}\,du=\frac{1}{10}\sin 32$$

\item[(iii)]

DIAGRAM

$$\int_{y=0}^3\!\int_{x=y^2}^{\frac{y^2}{3}+6}f(x,y)\,dxdy$$

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