integral from 0 to 1 of k(1-y)y^n

解答

\int_{0}^{1} k \cdot (1 - y) \cdot y^{n} d y

k (\frac{1}{n + 1} - \frac{1}{n + 2})

求解步骤

\int_{0}^{1} k (1 - y) y^{n} d y

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= k \cdot \int_{0}^{1} (1 - y) y^{n} d y

乘开

(1 - y) y^{n} : y^{n} - y^{1 + n}

= k \cdot \int_{0}^{1} y^{n} - y^{1 + n} d y

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= k (\int_{0}^{1} y^{n} d y - \int_{0}^{1} y^{1 + n} d y)

\int_{0}^{1} y^{n} d y = \frac{1}{n + 1}

\int_{0}^{1} y^{1 + n} d y = \frac{1}{n + 2}

= k (\frac{1}{n + 1} - \frac{1}{n + 2})

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