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