What is the integral from 0 to 2 of y(-y^2+4)/2 ?

The integral from 0 to 2 of y(-y^2+4)/2 is 2

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integral from 0 to 2 of y(-y^2+4)/2

\int_{0}^{2} y \frac{- y ^{2} + 4}{2} d y

2

Solution steps

\int_{0}^{2} y \frac{- y ^{2} + 4}{2} d y

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2} y (- y^{2} + 4) d y

Expand

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

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

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

\int_{0}^{2} y^{3} d y = 4

\int_{0}^{2} 4 y d y = 8

= \frac{1}{2} (- 4 + 8)

Simplify

\frac{1}{2} (- 4 + 8) : 2

= 2

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What is the integral from 0 to 2 of y(-y^2+4)/2 ?
The integral from 0 to 2 of y(-y^2+4)/2 is 2