Question Number 202167 by tri26112004 last updated on 22/Dec/23
$$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \underset{{x}} {\int}^{\mathrm{1}} {sin}\left({y}^{\mathrm{2}} \right){dydx}\:=\:¿ \\ $$
Answered by mnjuly1970 last updated on 22/Dec/23
$$\:{answer}:=\:{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Commented by tri26112004 last updated on 22/Dec/23
$${Solution}¿ \\ $$
Answered by witcher3 last updated on 22/Dec/23
![x≤y≤1 0≤x≤1⇔ 0≤x≤y..& 0≤y≤1 ∫_0 ^1 ∫_x ^1 sin(y^2 )dydx=∫_0 ^1 ∫_0 ^y sin(y^2 )dxdy =∫_0 ^1 ysin(y^2 )=−(1/2)[cos(y^2 )]_0 ^1 =(1/2)(1−cos(1)) cos(1)=cos(2.(1/2))=−2sin^2 ((1/2))+1 =sin^2 ((1/2))](https://www.tinkutara.com/question/Q202173.png)
$$\:\mathrm{x}\leqslant\mathrm{y}\leqslant\mathrm{1}\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}\Leftrightarrow\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{y}..\&\:\:\:\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{x}} ^{\mathrm{1}} \mathrm{sin}\left(\mathrm{y}^{\mathrm{2}} \right)\mathrm{dydx}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{y}} \mathrm{sin}\left(\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ysin}\left(\mathrm{y}^{\mathrm{2}} \right)=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left(\mathrm{y}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}\right)\right) \\ $$$$\mathrm{cos}\left(\mathrm{1}\right)=\mathrm{cos}\left(\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1} \\ $$$$=\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$