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Question
Evaluate: `int_0^1 "t"^2 sqrt(1 - "t") "dt"`
Solution
Let I = `int_0^1 "t"^2 sqrt(1 - "t") "dt"`
= `int_0^1 (1 -"t")^2 sqrt(1 - (1 - "t")) "dt"` ......`[∵ int_0^"a" "f"(x)"d"x = int_0^"a" "f"("a" - x)"d"x]`
= `int_0^1 (1 - 2"t" + "t"^2)sqrt("t") "dt"`
= `int_0^1("t"^(1/2) - 2"t"^(3/2) + "t"^(5/2))"dt"`
= `int_0^1 "t"^(1/2) "dt" - 2 int_0^1 "t"^(3/2) "dt" + int_0^1 "t"^(5/2) "dt"`
= `[("t"^(3/2))/(3/2)]_0^1 - 2[("t"^(5/2))/(5/2)]_0^1 + [("t"^(7/2))/(7/2)]_0^1`
= `2/3(1^(3/2) - 0) - 4/5(1^(5/2) - 0) + 2/7(1^(7/2) - 0)`
= `2/3 - 4/5 + 2/7`
= `(70 - 84 + 30)/105`
∴ I = `16/105`
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