What is the integral from 0 to 0.4 of 30y^2(1-y)^2 ?

The integral from 0 to 0.4 of 30y^2(1-y)^2 is 0.31744

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integral from 0 to 0.4 of 30y^2(1-y)^2

\int_{0}^{0.4} 30 y^{2} (1 - y)^{2} d y

0.31744

Solution steps

\int_{0}^{0.4} 30 y^{2} (1 - y)^{2} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 30 \cdot \int_{0}^{0.4} y^{2} (1 - y)^{2} d y

Expand

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

= 30 \cdot \int_{0}^{0.4} y^{2} - 2 y^{3} + y^{4} 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

= 30 (\int_{0}^{0.4} y^{2} d y - \int_{0}^{0.4} 2 y^{3} d y + \int_{0}^{0.4} y^{4} d y)

\int_{0}^{0.4} y^{2} d y = 0.02133 \dots

\int_{0}^{0.4} 2 y^{3} d y = 0.0128

\int_{0}^{0.4} y^{4} d y = 0.002048

= 30 (0.02133 \dots - 0.0128 + 0.002048)

Simplify

= 0.31744

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What is the integral from 0 to 0.4 of 30y^2(1-y)^2 ?
The integral from 0 to 0.4 of 30y^2(1-y)^2 is 0.31744