Question

CHANGE THE ORDER OF INTEGRA NON TO EVALUATE sin (x) dx dy

Answer 1

Solution

$\mathrm{\int _0^9\:\int _{\sqrt{y}}^3\:sin\left(\pi \:x^3\right)dx\:dy}$

$\mathrm{\mathrm{Change\:order\:of\:integration}}$

$\mathrm{\int _0^9\int _{\sqrt{y}}^3\sin \left(\pi x^3\right)dxdy}$

$\mathrm{\mathrm{Given:}\:0\le \:y\le \:9,\:\sqrt{y}\le \:x\le \:3}$

$\mathrm{\mathrm{For\:}\sqrt{y}\le \:x\le \:3\mathrm{\:plug\:in\:}y=0\quad \Rightarrow \:0\le \:x\le \:3}$

$\mathrm{\mathrm{Isolate}\:y\:\mathrm{for}\:\sqrt{y}\le \:x\le \:3\quad \Rightarrow \:0\le \:y\le \:x^2}$

$\mathrm{=\int _0^3\int _0^{x^2}\sin \left(\pi x^3\right)dydx}$

$\mathrm{\Rightarrow \int _0^{x^2}\sin \left(\pi x^3\right)dy=\left[\sin \left(x^3\pi \right)y\right]_0^{x^2}=\left[\sin \left(x^3\pi \right)y\right]^{x^2}_0=x^2\sin \left(\pi x^3\right)}$

$\Rightarrow =\int _0^3x^2\sin \left(\pi x^3\right)dx$

$\mathrm{\mathrm{Apply\:u-substitution:}\:u=\pi x^3}$

$\mathrm{\frac{du}{dx}=3\pi x^2}$

$\mathrm{\Rightarrow \:du=3\pi x^2dx}$

$\mathrm{\Rightarrow \:dx=\frac{1}{3\pi x^2}du}$

$\mathrm{=\int \:x^2\sin \left(u\right)\frac{1}{3\pi x^2}du}$

$\mathrm{=\int \frac{\sin \left(u\right)}{3\pi }du}$

Adjust integral boundaries :

$\mathrm{x=0\quad \Rightarrow \:u=0}$

$\mathrm{\:x=3\quad \Rightarrow \:u=27\pi }$

$\mathrm{=\int _0^{27\pi }\frac{\sin \left(u\right)}{3\pi }du=\frac{1}{3\pi }\cdot \int _0^{27\pi }\sin \left(u\right)du}$

$\mathrm{\mathrm{Use\:the\:common\:integral}:\quad \int \sin \left(u\right)du=-\cos \left(u\right)}$

$\mathrm{\mathrm{=\frac{1}{3\pi }\left[-\cos \left(u\right)\right]^{27\pi }_0}=\left[-\cos \left(u\right)\right]^{27\pi }_0=2}$

$\mathrm{\mathrm{=\frac{1}{3\pi }\cdot \:2}=\frac{2}{3\pi }}$

$\mathrm{=\frac{2}{3\pi }}$