integral from 0 to H of z^2(1-z)^2

解

\int_{0}^{H} z^{2} (1 - z)^{2} d z

\frac{H ^{3}}{3} - \frac{H ^{4}}{2} + \frac{H ^{5}}{5}

解答ステップ

\int_{0}^{H} z^{2} (1 - z)^{2} d z

拡張

z^{2} (1 - z)^{2} : z^{2} - 2 z^{3} + z^{4}

= \int_{0}^{H} z^{2} - 2 z^{3} + z^{4} d z

総和規則を適用する:

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

= \int_{0}^{H} z^{2} d z - \int_{0}^{H} 2 z^{3} d z + \int_{0}^{H} z^{4} d z

\int_{0}^{H} z^{2} d z = \frac{H ^{3}}{3}

\int_{0}^{H} 2 z^{3} d z = \frac{H ^{4}}{2}

\int_{0}^{H} z^{4} d z = \frac{H ^{5}}{5}

= \frac{H ^{3}}{3} - \frac{H ^{4}}{2} + \frac{H ^{5}}{5}