int_0^1 x(1-x)^n dx = . . . . . . . | vedclass

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(C) To evaluate the integral $I = \int_0^1 x(1-x)^n dx$,we use the property of definite integrals: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Applying t

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$\frac{1}{n^2+3n+2}$

Vedclass · Answer

(C) To evaluate the integral $I = \int_0^1 x(1-x)^n dx$,we use the property of definite integrals: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Applying this property,we get: $I = \int_0^1 (1-x)(1-(1-x))^n dx$ $I = \int_0^1 (1-x)x^n dx$ $I = \int_0^1 (x^n - x^{n+1}) dx$ Now,integrate term by term: $I = [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}]_0^1$ $I = \frac{1}{n+1} - \frac{1}{n+2}$ $I = \frac{(n+2) - (n+1)}{(n+1)(n+2)}$ $I = \frac{1}{n^2 + 3n + 2}$ Thus,the correct option is $C$.

$\int_0^1 x(1-x)^n dx = $ . . . . . . .

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