Question Number 213074 by Spillover last updated on 29/Oct/24
Answered by MrGaster last updated on 29/Oct/24
![=∫_0 ^1 ∫_0 ^(1/2) [−((cos(x+y^2 +z^3 ))/(3z^2 ))]_0 ^(1/3) dydx =∫_0 ^1 ∫_0 ^(1/2) (−((cos(x+y^2 +(1/(27))))/(3∙(1/9)))+((cos(x+y^2 ))/(3∙0^2 )))dydx =∫_0 ^1 ∫_0 ^(1/2) (3 cos(x+y^2 )−3 cos(x+y^2 +(1/(27))))dydx =∫_0 ^1 [3 sin(x+y^2 )−3 sin(x+y^2 +(1/(27)))]_0 ^(1/2) dx =∫_0 ^1 (3 sin(x+(1/4))−3 sin(x+(1/4)+(1/(27)))−3 sin(x)+3 sin(x+(1/(27))))dx =[−3 cos(x+(1/4))+3 cos(x+(1/4)+(1/(27)))+ 3 cos(x)−3 cos(x+(1/(27)))]_0 ^1 =− 3 cos(1+(1/4))+3 cos(1+(1/4)+(1/(27)))+3 cos(1)−3 cos(1+(1/(27)))+3cos((1/4))− 3 cos((1/4)+(1/(27)))−3 cos(0)+3 cos((1/(27)))](https://www.tinkutara.com/question/Q213076.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)}{\mathrm{3}{z}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right)}{\mathrm{3}\centerdot\frac{\mathrm{1}}{\mathrm{9}}}+\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} \right)}{\mathrm{3}\centerdot\mathrm{0}^{\mathrm{2}} }\right){dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{3}\:\mathrm{cos}\left({x}+{y}^{\mathrm{2}} \right)−\mathrm{3}\:\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right)\right){dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\mathrm{3}\:\mathrm{sin}\left({x}+{y}^{\mathrm{2}} \right)−\mathrm{3}\:\mathrm{sin}\left({x}+{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}\right)\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\right]}_{\mathrm{0}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}\:\mathrm{sin}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{3}\:\mathrm{sin}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)−\mathrm{3}\:\mathrm{sin}\left({x}\right)+\mathrm{3}\:\mathrm{sin}\left({x}+\frac{\mathrm{1}}{\mathrm{27}}\right)\right){dx} \\ $$$$=\left[−\mathrm{3}\:\mathrm{cos}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3}\:\mathrm{cos}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\:\mathrm{3}\:\mathrm{cos}\left({x}\right)−\mathrm{3}\:\mathrm{cos}\left({x}+\frac{\mathrm{1}}{\mathrm{27}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\:\mathrm{3}\:\mathrm{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{3}\:\mathrm{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{3}\:\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{3}\:\mathrm{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{27}}\right)+\mathrm{3cos}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\:\mathrm{3}\:\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}\right)−\mathrm{3}\:\mathrm{cos}\left(\mathrm{0}\right)+\mathrm{3}\:\mathrm{cos}\left(\frac{\mathrm{1}}{\mathrm{27}}\right) \\ $$
Commented by Frix last updated on 04/Nov/24
$$\mathrm{No}. \\ $$