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Question
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Sum
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Solution
Le I = `int_0^1 log x/((1 - x)^(3/4)) "d"x`
Using the property
`int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x`
I = `int_0^1 ((1 - x))/[1 - (1 - x)]^(3/4) "d"x`
= `int_0^1 ((1 - x))/(x)^(3/4) "d"x`
= `int_0^1 (1 - x) (x)^((-3)/4) "d"x`
= `int_^1 (x^((-3)/4) - x^(1 - 3/4)) "d"x`
= `int_^1 (x^((-3)/4) - x^(1/4)) "d"x`
= `[(x^(-3/4 + 1))/(((-3)/4 + 1))]_0^1 - [x^(1/4 + 1)/((1/4 + 1))]_0^1`
= `[x^(1/4)/((1/4))]_0^1 - [x^(5/4)/((5/4))]_0^1`
= `4[x^(1/4)]_0^1 - 4/5 [x^(5/4)]_0^1`
= `4[1 - 0] - 4/5 [1 - 0]`
= `4 - 4/5`
= `(20 - 4)/5`
∴ I = `16/5`
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Definite Integrals
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