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

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(B) 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 have: $I = \int_{0}^{1} (1 - x)(1

Vedclass · Accepted Answer

$\frac{1}{42}$

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(B) 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 have: $I = \int_{0}^{1} (1 - x)(1 - (1 - x))^{5} dx$ $I = \int_{0}^{1} (1 - x)(x)^{5} dx$ $I = \int_{0}^{1} (x^{5} - x^{6}) dx$ Integrating term by term: $I = \left[ \frac{x^{6}}{6} - \frac{x^{7}}{7} \right]_{0}^{1}$ $I = \left( \frac{1}{6} - \frac{1}{7} \right) - (0 - 0)$ $I = \frac{7 - 6}{42} = \frac{1}{42}$.

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

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