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Question Number 144788 by nonh1 last updated on 29/Jun/21

Answered by gsk2684 last updated on 30/Jun/21

$$\underset{\mathrm{x}=0}{\stackrel{1}{\int }}\left(\underset{\mathrm{y}=-3}{\stackrel{3}{\int }}(4-{\mathrm{y}}^2)\mathrm{dy}\right)\mathrm{dx}$$$$\underset{0}{\stackrel{1}{\int }}{[4\mathrm{y}-\frac{{\mathrm{y}}^3}{3}]}_{-3}^3\mathrm{dx}$$$$\underset{0}{\stackrel{1}{\int }}(12-9)-(-12+9)\mathrm{dx}$$$$\underset{0}{\stackrel{1}{\int }}6\mathrm{dx}=6{\left[\mathrm{x}\right]}_0^1=6(1-0)=6$$

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