Evaluate ∫0ax2(a-x)32 dx - Mathematics and Statistics

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Sum

Evaluate `int_0^"a" x^2 ("a" - x)^(3/2)  "d"x`

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Solution

Let I = `int_0^"a" x^2 ("a" - x)^(3/2)  "d"x`

= `int_0^"a" ("a" - x)^2 ["a" - ("a" - x)]^(3/2)  "d"x`    ......`[because int_0^"a" "f"(x)  "d"x = int_0^"a" "f"("a" - x) "d"x]`

= `int_0^"a"("a"^2 - 2"a"x + x^2)x^(3/2)  "d"x`

= `int_0^"a"("a"^2x^(3/2) - 2"a"x^(5/2) + x^(7/2))"d"x`

= `"a"^2 int_0^"a" x^(3/2) "d"x - 2"a" int_0^"a" x^(5/2)  "d"x + int_0^"a" x^(7/2)  "d"x`

= `"a"^2[(x^(5/2))/(5/2)]_0^"a" - 2"a"[(x^(7/2))/(7/2)]_0^"a" + [(x^(9/2))/(9/2)]_0^"a"`

= `(2"a"^2)/5 [("a")^(5/2) - 0] - (4"a")/7 [("a")^(7/2) - 0] + 2/9 [("a")^(9/2) - 0]`

= `2/5"a"^(9/2) - 4/7"a"^(9/2) + 2/9"a"^(9/2)`

= `(2/5 - 4/7 + 2/9)"a"^(9/2)`

= `((126 - 180 + 70)/315)"a"^(9/2)`

∴ I = `16/315"a"^(9/2)`

Concept: Fundamental Theorem of Integral Calculus
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Chapter 1.6: Definite Integration - Q.5

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