Question Number 205935 by EJJDJX last updated on 03/Apr/24
$$\int\underset{{D}} {\int}\left(\mathrm{4}{y}^{\mathrm{2}} {sin}\left({xy}\right)\right){dxdy}\:\:=\:??? \\ $$$${D}:\:\:\:\:\:\:\:{x}={y}\:\:\:\:\:\:{x}=\mathrm{0}\:\:\:\:\:\:\:{y}=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant{y}\:\:\:\:\:\:\:\mathrm{0}\leqslant{y}\leqslant\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$
Answered by Berbere last updated on 03/Apr/24
![∫_0 ^(√(π/2)) ∫_0 ^y 4y^2 sin(xy)dxdy =∫_0 ^(√(π/2)) −4y[cos(xy)]_0 ^y dy =∫_0 ^(√(π/2)) −4y[cos(y^2 )−1)dy =−2∫_0 ^(√(π/2)) (cos(y^2 )−1)dy^2 =2∫_0 ^(√(π/2)) (sin(y^2 )−y^2 ) =2(1−(π/2))](https://www.tinkutara.com/question/Q205942.png)
$$\int_{\mathrm{0}} ^{\sqrt{\frac{\pi}{\mathrm{2}}}} \int_{\mathrm{0}} ^{{y}} \mathrm{4}{y}^{\mathrm{2}} {sin}\left({xy}\right){dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\frac{\pi}{\mathrm{2}}}} −\mathrm{4}{y}\left[{cos}\left({xy}\right)\right]_{\mathrm{0}} ^{{y}} {dy} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\frac{\pi}{\mathrm{2}}}} −\mathrm{4}{y}\left[{cos}\left({y}^{\mathrm{2}} \right)−\mathrm{1}\right){dy} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\frac{\pi}{\mathrm{2}}}} \left({cos}\left({y}^{\mathrm{2}} \right)−\mathrm{1}\right){dy}^{\mathrm{2}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\frac{\pi}{\mathrm{2}}}} \left({sin}\left({y}^{\mathrm{2}} \right)−{y}^{\mathrm{2}} \right) \\ $$$$=\mathrm{2}\left(\mathrm{1}−\frac{\pi}{\mathrm{2}}\right) \\ $$