What is the integral from 0 to H of z^2(1-z)^2 ?

The integral from 0 to H of z^2(1-z)^2 is (H^3)/3-(H^4)/2+(H^5)/5

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\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}

What is the integral from 0 to H of z^2(1-z)^2 ?
The integral from 0 to H of z^2(1-z)^2 is (H^3)/3-(H^4)/2+(H^5)/5