Question Number 144788 by nonh1 last updated on 29/Jun/21
Answered by gsk2684 last updated on 30/Jun/21
![∫_(x=0) ^1 (∫_(y=−3) ^3 (4−y^2 )dy) dx ∫_0 ^1 [4y−(y^3 /3)]_(−3) ^3 dx ∫_0 ^1 (12−9)−(−12+9)dx ∫_0 ^1 6dx=6[x]_0 ^1 =6(1−0)=6](https://www.tinkutara.com/question/Q144921.png)
$$\underset{\mathrm{x}=\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\underset{\mathrm{y}=−\mathrm{3}} {\overset{\mathrm{3}} {\int}}\left(\mathrm{4}−\mathrm{y}^{\mathrm{2}} \right)\mathrm{dy}\right)\:\mathrm{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\mathrm{4y}−\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{3}} ^{\mathrm{3}} \mathrm{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{12}−\mathrm{9}\right)−\left(−\mathrm{12}+\mathrm{9}\right)\mathrm{dx} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{6dx}=\mathrm{6}\left[\mathrm{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{6}\left(\mathrm{1}−\mathrm{0}\right)=\mathrm{6} \\ $$