Let beta(m, n) = int_0^1 x^{m-1}(1-x)^{n-1} d | vedclass

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(D) Given the integral $I = \int_0^1 (1-x^{10})^{20} dx$. Let $x^{10} = t$,then $x = t^{1/10}$. Differentiating both sides,$dx = \frac{1}{10} t^{1/10

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$2120$

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(D) Given the integral $I = \int_0^1 (1-x^{10})^{20} dx$. Let $x^{10} = t$,then $x = t^{1/10}$. Differentiating both sides,$dx = \frac{1}{10} t^{1/10 - 1} dt = \frac{1}{10} t^{-9/10} dt$. Substituting these into the integral: $I = \int_0^1 (1-t)^{20} \cdot \frac{1}{10} t^{-9/10} dt = \frac{1}{10} \int_0^1 t^{-9/10} (1-t)^{20} dt$. Comparing this with the definition $\beta(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$,we identify: $m-1 = -9/10 \implies m = 1/10$ and $n-1 = 20 \implies n = 21$. Thus,$I = \frac{1}{10} \beta(1/10, 21)$. Comparing with $a 	imes \beta(b, c)$,we get $a = 1/10$,$b = 1/10$,and $c = 21$. Finally,$100(a+b+c) = 100(1/10 + 1/10 + 21) = 100(0.2 + 21) = 100(21.2) = 2120$.

Let $\beta(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$,where $m, n > 0$. If $\int_0^1 (1-x^{10})^{20} dx = a \times \beta(b, c)$,then $100(a+b+c)$ equals:

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