Q. 12 - 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.

${\int }_0^{\sqrt{\mathrm{\pi }}}{\int }_x^{\sqrt{\mathrm{\pi }}}\sin y^2 dydx$

Step-by-Step Solution

Answer

${\int }_0^{\sqrt{\mathrm{\pi }}}{\int }_x^{\sqrt{\mathrm{\pi }}}\sin y^2 dydx=1$

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.

${\int }_0^{\sqrt{\mathrm{\pi }}}{\int }_x^{\sqrt{\mathrm{\pi }}}\sin y^2 dydx\Rightarrow {\int }_0^{\sqrt{\mathrm{\pi }}}{\int }_0^y\sin y^2 dydx$

Step 3: Evaluate the integral

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