Weird Integral problem?
Given Integral (limits 0-1) [f(x)]^4 dx = 1/5 and Integral (limits 0-1) [f(x)]^2 dx = 1/3 , find Integral (limits 0-1) [f(x)]^2(2[f(x)]^(2)-3)dx ?
step by step please, I am lost.
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Given Integral (limits 0-1) [f(x)]^4 dx = 1/5 and Integral (limits 0-1) [f(x)]^2 dx = 1/3 , find Integral (limits 0-1) [f(x)]^2(2[f(x)]^(2)-3)dx ?
step by step please, I am lost.
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Verified answer
∫(f(x))^2 * (2(f(x))^2 - 3) dx from 0 to 1
= ∫(2(f(x))^4 - 3(f(x))^2) dx from 0 to 1
= 2(1/5) - 3(1/3) = -3/5