If I ( m , n ) = 0 1 x m – 1 ( 1 – x ) n – 1 d x , m, n 0, then I (9, 14) + I (10, 13) is [2025]

Question

If $I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x$ , m, n > 0, then $I$ (9, 14) + $I$ (10, 13) is [2025]

Smart Achievers · Accepted Answer

(3)

$I (m, n) = \int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} d x, m, n > 0$

Let $x = \sin^{2} θ \Rightarrow d x = 2 sinθ cosθ dθ$

$∴ I (m, n) = \int_{0}^{π / 2} \sin^{2 m - 2} θ \times {(1 - \sin^{2} θ)}^{n - 1} \times 2 sinθcosθ d θ$

$= 2 \int_{0}^{π / 2} \sin^{2 m - 1} θ \times \cos^{2 n - 1} θ d θ$

$∴ I (9, 14) + I (10, 13) = 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{27} θ dθ + 2 \int_{0}^{π / 2} \sin^{19} θ \cos^{25} θ dθ$

$= 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{25} θ (\cos^{2} θ + \sin^{2} θ) d θ$

$= 2 \int_{0}^{π / 2} \sin^{17} θ \cos^{25} θ dθ = I (9, 13)$ .

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