Q. 11 - Calculus

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.

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

Step-by-Step Solution

Answer

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

Step 1: Draw the region.

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

Step 2: Reversing the order of integration.

From the above diagram, reversing the order of integration.

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

Step 3: Evaluate the integral

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