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

The integral from 0 to 2 of y(2y^2-(y^2)/2) is 6

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integral from 0 to 2 of y(2y^2-(y^2)/2)

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

6

Solution steps

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

Simplify

2 y^{2} - \frac{y ^{2}}{2} : \frac{3 y ^{2}}{2}

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

Take the constant out:

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

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

Simplify

y y^{2} : y^{3}

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

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{0}^{2}

= \frac{3}{2} [\frac{y ^{4}}{4}]_{0}^{2}

Compute the boundaries:

4

= \frac{3}{2} \cdot 4

Simplify

= 6

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What is the integral from 0 to 2 of y(2y^2-(y^2)/2) ?
The integral from 0 to 2 of y(2y^2-(y^2)/2) is 6