Question Number 199942 by cortano12 last updated on 11/Nov/23
Answered by MM42 last updated on 11/Nov/23
![1−(√x)=u^2 ⇒x=(1−u^2 )^2 =x⇒dx=−4u(1−u^2 )du ⇒I=4∫_0 ^1 (u^2 −u^4 )du=4((1/3)u^3 −(1/5)u^5 )]_0 ^1 =4((1/3)−(1/5))= (8/(15)) ✓](Q199944.png)
$$\mathrm{1}−\sqrt{{x}}={u}^{\mathrm{2}} \Rightarrow{x}=\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\Rightarrow{dx}=−\mathrm{4}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right){du} \\ $$$$\left.\Rightarrow{I}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left({u}^{\mathrm{2}} −{u}^{\mathrm{4}} \right){du}=\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\right)=\:\frac{\mathrm{8}}{\mathrm{15}}\:\checkmark \\ $$
Commented by cortano12 last updated on 11/Nov/23
$$\mathrm{yes} \\ $$