What is the integral from 0 to 1 of y^2(1-y^{3/2}) ?

The integral from 0 to 1 of y^2(1-y^{3/2}) is 1/9

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integral from 0 to 1 of y^2(1-y^{3/2})

\int_{0}^{1} y^{2} (1 - y^{\frac{3}{2}}) d y

\frac{1}{9}

Decimal

0.11111 \dots

Solution steps

\int_{0}^{1} y^{2} (1 - y^{\frac{3}{2}}) d y

Expand

y^{2} (1 - y^{\frac{3}{2}}) : y^{2} - y^{\frac{7}{2}}

= \int_{0}^{1} y^{2} - y^{\frac{7}{2}} d y

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}^{1} y^{2} d y - \int_{0}^{1} y^{\frac{7}{2}} d y

\int_{0}^{1} y^{2} d y = \frac{1}{3}

\int_{0}^{1} y^{\frac{7}{2}} d y = \frac{2}{9}

= \frac{1}{3} - \frac{2}{9}

Simplify

\frac{1}{3} - \frac{2}{9} : \frac{1}{9}

= \frac{1}{9}

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What is the integral from 0 to 1 of y^2(1-y^{3/2}) ?
The integral from 0 to 1 of y^2(1-y^{3/2}) is 1/9