Evaluate the following integrals : $\int_0^1 x(1-x)^5 d x$
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Answer
We use the property $ \begin{aligned} \int_0^a f(x) d x=\int_0^a f(a-x) d x \\ \therefore \int_0^1 x(1-x)^5 d x & =\int_0^1(1-x)(1-1+x)^5 d x \\ =\int_0^1(1-x) x^5 d x=\int_0^1\left(x^5-x^6\right) d x \\ =\int_0^1 x^5 d x-\int_0^1 x^6 d x \\ =\left[\frac{x^6}{6}\right]_0^1-\left[\frac{x^7}{7}\right]_0^1 \\ =\frac{1}{6}(1-0)-\frac{1}{7}(1-0) \\ =\frac{1}{6}-\frac{1}{7}=\frac{1}{42} . \end{aligned} $
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