Question

Reversing the order of integration: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals by reversing the order of integration.

${\int }_0^4{\int }_{\sqrt{x}}^{2}y\cos xdydx$

Short answer

${\int }_0^4{\int }_{\sqrt{x}}^{2}y\cos xdydx=\frac{1-\cos 4}{2}$

Step-by-step solution

Draw the region.

The region determined by the limits of the given iterated integral is shown below,

Reversing the order of integration.

From the above diagram, reversing the order of integration.

${\int }_0^4{\int }_{\sqrt{x}}^{2}y\cos xdydx\Rightarrow {\int }_0^{y^2}{\int }_0^{2}y\cos xdydx$

Evaluate the integral

$$\begin{gathered} I={\int }_0^{y^2}{\int }_0^{2}y\cos xdydx \\ I={\int }_0^{2}y\left[{\int }_0^{y^2}\cos xdx\right]dy \\ I={\int }_0^{2}y{\left[\sin x\right]}_0^{y^2}dy \\ I={\int }_0^{2}y\sin y^{2}dy \\ I=\frac{1}{2}{\left[-\cos y^2\right]}_0^2 \\ I=\frac{1}{2}\left[-\cos 4+1\right] \\ I=\frac{1-\cos 4}{2} \end{gathered}$$