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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)) ✓

1−x=u2⇒x=(1−u2)2=x⇒dx=−4u(1−u2)du⇒I=4∫01(u2−u4)du=4(13u3−15u5)]01=4(13−15)=815✓

Commented by cortano12 last updated on 11/Nov/23

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