integral of k(1-y)y^n

解答

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

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

求解步骤

\int 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 (1 - y) y^{n} d y

乘开

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

= k \cdot \int 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 y^{n} d y - \int y^{1 + n} d y)

\int y^{n} d y = \frac{y ^{n + 1}}{n + 1}

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

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

解答补常数

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