int_{0}^{1} x(1-x)^{5} dx = | vedclass

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(C) Let $I = \int_{0}^{1} x(1-x)^{5} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{1} (1-x)(1-(1-x))^

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$\frac{1}{42}$

Vedclass · Answer

(C) Let $I = \int_{0}^{1} x(1-x)^{5} dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{1} (1-x)(1-(1-x))^{5} dx$ $I = \int_{0}^{1} (1-x)x^{5} dx$ $I = \int_{0}^{1} (x^{5} - x^{6}) dx$ $I = \left[ \frac{x^{6}}{6} - \frac{x^{7}}{7} \right]_{0}^{1}$ $I = \left( \frac{1}{6} - \frac{1}{7} \right) - (0 - 0)$ $I = \frac{7-6}{42} = \frac{1}{42}$.

$\int_{0}^{1} x(1-x)^{5} dx =$

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