Let $\beta(m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} dx, m, n>0$. If $\int_{0}^{1}(1-x^{10})^{20} dx=a \times \beta(b, c)$, then $100(a+b+c)$ equals ____ .

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Definite Integrals

Let

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

. If

\int_{0}^{1} (1 - x^{10})^{20} d x = a \times β (b, c)

, then

100 (a + b + c)

equals ____ .

Answer: (B)

Substitute

x^{10} = u, x = u^{1/10}, d x = \frac{1}{10} u^{- 9/10} d u

\int_{0}^{1} (1 - x^{10})^{20} d x = \frac{1}{10} \int_{0}^{1} u^{1/10 - 1} (1 - u)^{20} d u = \frac{1}{10} β (\frac{1}{10}, 21)

a = \frac{1}{10}, b = \frac{1}{10}, c = 21

100 (a + b + c) = 100 (\frac{1}{10} + \frac{1}{10} + 21) = 100 (21.2) = 2120

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option

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