Integration Questions

Question Number 150678 by krauss last updated on 14/Aug/21

$$\\$$ $$\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}−{x}^{\mathrm{2}} } \left(\mathrm{3}−{x}^{\mathrm{2}} −{y}\right){dy}\:{dx} \\$$

Answered by Ar Brandon last updated on 14/Aug/21

$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{3}−{x}^{\mathrm{2}} } \left(\mathrm{3}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dydx}=\int_{\mathrm{0}} ^{\mathrm{2}} \left[\mathrm{3}{y}−{x}^{\mathrm{2}} {y}−\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{3}−{x}^{\mathrm{2}} } {dx} \\$$ $$\:\:=\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{3}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} \left(\mathrm{3}−{x}^{\mathrm{2}} \right)−\frac{\left(\mathrm{3}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{3}}\right){dx} \\$$ $$\mathrm{You}\:\mathrm{may}\:\mathrm{proceed}\:\mathrm{by}\:\mathrm{expanding}\:\mathrm{and}\:\mathrm{integrating}. \\$$

Answered by tabata last updated on 15/Aug/21

$$=\int_{\mathrm{0}} ^{\:\mathrm{2}} \left(\mathrm{3}{y}−{x}^{\mathrm{2}} {y}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)_{\mathrm{0}} ^{\mathrm{3}−{x}^{\mathrm{2}} } {dx}\:=\int_{\mathrm{0}} ^{\:\mathrm{2}} \left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right){dx} \\$$ $$\\$$ $$=\left(\frac{\mathrm{9}}{\mathrm{2}}{x}−{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{10}}{x}^{\mathrm{5}} \right)_{\mathrm{0}} ^{\mathrm{2}} =\left(\mathrm{9}−\mathrm{8}+\frac{\mathrm{32}}{\mathrm{10}}\right)=\left(\mathrm{1}+\frac{\mathrm{32}}{\mathrm{10}}\right)=\frac{\mathrm{42}}{\mathrm{10}}=\frac{\mathrm{21}}{\mathrm{5}} \\$$