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प्रश्न
Evaluate `int_0^1 (x^2 + 3x + 2)/sqrt(x) "d"x`
उत्तर
Let I = `int_0^1 (x^2 + 3x + 2)/sqrt(x) "d"x`
= `int_0^1 ((x^2 + 3x + 2)/x^(1/2)) "d"x`
= `int_0^1 (x^2/x^(1/2) + (3x)/(x^(1/2)) + 2/(x^(1/2))) "d"x`
= `int_0^1 (x^(3/2) + 3x^(1/2) + 2x^(-1/2)) "d"x`
= `int_0^1 x^(3/2) "d"x + 3int_0^1 x^(1/2) "d"x + 2int_0^1 x^(-1/2) "d"x`
= `[(x^(5/2))/(5/2)]_0^1 + 3[(x^(3/2))/(3/2)]_0^1 + 2[(x^(1/2))/(1/2)]_0^1`
= `2/5(1 - 0) + 3 xx 2/3(1 - 0) + 2 xx 2(1 - 0)`
= `2/5 + 2 + 4`
∴ I = `32/5`
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