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

The integral from (y^2)/4 to (y+4)/2 of 2y is 2((y^2)/8+y+2-(y^4)/(32))

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

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

2 (\frac{y ^{2}}{8} + y + 2 - \frac{y ^{4}}{32})

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{\frac{y ^{2}}{4}}^{\frac{y + 4}{2}} y d y

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{8} y^{2} + y + 2 - \frac{y ^{4}}{32}

= 2 (\frac{1}{8} y^{2} + y + 2 - \frac{y ^{4}}{32})

Simplify

= 2 (\frac{y ^{2}}{8} + y + 2 - \frac{y ^{4}}{32})

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What is the integral from (y^2)/4 to (y+4)/2 of 2y ?
The integral from (y^2)/4 to (y+4)/2 of 2y is 2((y^2)/8+y+2-(y^4)/(32))