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Question
Evaluate `int _0^l "x" (1 - "x")^(3/2) "dx"`
Solution
Let I = `int _0^1 "x" (1 - "x")^(3/2) "dx"`
By using the property
`int_0^a "f(x) dx" = int_0^a "f (a - x) dx"`
I = `int_0^1 (1 - "x") [1-(1 - "x")]^(3/2) "dx"`
`= int _0^1 (1 - "x") . "x"^(3/2) "dx"`
`= int _0^1 ("x"^(3/2) - "x"^(5/2)) "dx"`
`= int _0^1 "x"^(3/2) "dx" - int _0^1 "x"^(5/2) "dx"`
`= [("x"^(5/2))/(5/2)]_0^1 - [("x")^(7/2)/(7/2)]_0^1`
`= 2/5 [1 - 0] - 2/7 [1 - 0]`
`= 2/5 - 2/7 = 4/35`
`therefore "I" = 4/35`
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