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प्रश्न
Evaluate the following integrals : `int_0^"a" x^2("a" - x)^(3/2)*dx`
उत्तर
Let I = `int_0^"a" x^2("a" - x)^(3/2)*dx`
= `int_0^"a"("a" -x)^2 ["a" - ("a" - x)]^(3/2)*dx ...[because int_0^"a" f(x)*dx = int_0^"a" f("a" - x)*dx]`
= `int_0^"a" ("a"^2 - 2"a"x + x^2)x^(3/2)*dx`
= `int_0^"a"("a"^2x^(3/2) - 2"a"x^(5/2) + x^(7/2))*dx`
= `"a"^2 int_0^"a" x^(3/2)*dx - 2"a" int_0^"a" x^(5/2)*dx + int_0^"a"x^(7/2)*dx`
= `"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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