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\int_{0}^{1} x {(1 - x)}^{n} dx Find the \int x {(1 - x)}^{n} dx Let 1 - x = t - dx = dt \int x {(1 - x)}^{n} dx = - \int (1 - t) t^{n} dt = - \int (t^{n} - t^{n + 1}) dt = \int (t^{n + 1} - t^{n}) dt = [\frac{t^{n + 2}}{n + 2} - \frac{t^{n + 1}}{n + 1}] = [\frac{{(1 - x)}^{n + 2}}{n + 2} - \frac{{(1 - x)}^{n + 1}}{n + 1}] Now, \int_{0}^{1} x {(1 - x)}^{n} dx = {[\frac{{(1 - x)}^{n + 2}}{n + 2} - \frac{{(1 - x)}^{n + 1}}{n + 1}]}_{0}^{1} = [0 - (\frac{1}{n + 2} - \frac{1}{n + 1})] = \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{n + 2 - n - 1}{(n + 1) (n + 2)} = \frac{1}{(n + 1) (n + 2)} ANS . . .

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