What is the integral from 0 to 2-y of 3/4 y ?

The integral from 0 to 2-y of 3/4 y is (3(-y+2)^2)/8

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integral from 0 to 2-y of 3/4 y

\int_{0}^{2 - y} \frac{3}{4} y d y

\frac{3 ( - y + 2 ) ^{2}}{8}

Solution steps

\int_{0}^{2 - y} \frac{3}{4} y d y

Take the constant out:

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

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

Apply the Power Rule

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

Compute the boundaries:

\frac{1}{2} (- y + 2)^{2}

= \frac{3}{4} \cdot \frac{1}{2} (- y + 2)^{2}

Simplify

= \frac{3 ( - y + 2 ) ^{2}}{8}

What is the integral from 0 to 2-y of 3/4 y ?
The integral from 0 to 2-y of 3/4 y is (3(-y+2)^2)/8