What is the integral from 0 to 1 of (2-y)^2sqrt(y) ?

The integral from 0 to 1 of (2-y)^2sqrt(y) is 142/105

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integral from 0 to 1 of (2-y)^2sqrt(y)

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

\frac{142}{105}

Decimal Notation

1.35238 \dots

Solution steps

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

Expand

(2 - y)^{2} y : 4 y - 4 y^{\frac{3}{2}} + y^{\frac{5}{2}}

= \int_{0}^{1} 4 y - 4 y^{\frac{3}{2}} + y^{\frac{5}{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} 4 y d y - \int_{0}^{1} 4 y^{\frac{3}{2}} d y + \int_{0}^{1} y^{\frac{5}{2}} d y

\int_{0}^{1} 4 y d y = \frac{8}{3}

\int_{0}^{1} 4 y^{\frac{3}{2}} d y = \frac{8}{5}

\int_{0}^{1} y^{\frac{5}{2}} d y = \frac{2}{7}

= \frac{8}{3} - \frac{8}{5} + \frac{2}{7}

Simplify

\frac{8}{3} - \frac{8}{5} + \frac{2}{7} : \frac{142}{105}

= \frac{142}{105}

\int_{0}^{2} 4 x (x^{2} + 3)^{3} d x

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\int_{0}^{2} \frac{x ^{6}}{1 + x ^{2}} d x

What is the integral from 0 to 1 of (2-y)^2sqrt(y) ?
The integral from 0 to 1 of (2-y)^2sqrt(y) is 142/105