In I(m,n)=∫01 xm−1(1−x)n−1dx,m,n>0, then I(9, 14) + I(10, 13) is

I(m,m)=∫01 xm−1(1−x)n−1dxLet x=sin2⁡θ dx=2sin⁡θcos⁡θdθI(m,n)=2∫0π/2 (sin⁡θ)2m−1(cos⁡θ)2n−1dθI(9,14)+I(10,13)=2∫0π/2 (sin⁡θ)17(cos⁡θ)27dθ+2∫0π/2 (sin⁡θ)19(cos⁡θ)25dθ=2∫0π/2 (sin⁡θ)17(cos⁡θ)25dθ=I(9,13)

In I(m,n)=∫01 xm−1(1−x)n−1dx,m,n>0, then I(9, 14) + I(10, 13) is

I(m,m)=∫01 xm−1(1−x)n−1dxLet x=sin2⁡θ dx=2sin⁡θcos⁡θdθI(m,n)=2∫0π/2 (sin⁡θ)2m−1(cos⁡θ)2n−1dθI(9,14)+I(10,13)=2∫0π/2 (sin⁡θ)17(cos⁡θ)27dθ+2∫0π/2 (sin⁡θ)19(cos⁡θ)25dθ=2∫0π/2 (sin⁡θ)17(cos⁡θ)25dθ=I(9,13)

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In $I (m, n) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} dx, m, n > 0$ , then I(9, 14) + I(10, 13) is

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I(9, 13)

I(9, 1)

I(19, 27)

I(1, 13)

answer is D.

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Detailed Solution

$I (m, m) = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} dx$
Let $x = \sin^{2} θ dx = 2 \sin θcos θdθ$
$\begin{array}{l} I (m, n) = 2 \int_{0}^{π / 2} (\sin θ)^{2 m - 1} (\cos θ)^{2 n - 1} dθ \\ I (9, 14) + I (10, 13) = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{27} dθ + 2 \int_{0}^{π / 2} (\sin θ)^{19} (\cos θ)^{25} dθ \\ = 2 \int_{0}^{π / 2} (\sin θ)^{17} (\cos θ)^{25} dθ \\ = I (9, 13) \end{array}$