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

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(B) ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $\int_0^1 x(1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx$ $= \int_0^1 (1-x

Vedclass · Accepted Answer

$\frac{1}{(n+1)(n+2)}$

Vedclass · Answer

(B) ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,આપણને મળે છે: $\int_0^1 x(1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx$ $= \int_0^1 (1-x)x^n dx$ $= \int_0^1 (x^n - x^{n+1}) dx$ $= \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_0^1$ $= \left( \frac{1}{n+1} - \frac{1}{n+2} \right) - (0 - 0)$ $= \frac{(n+2) - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)}$

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

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