What is the integral of (4-y)/2 ?

The integral of (4-y)/2 is 1/2 (4y-(y^2)/2)+C

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integral of (4-y)/2

\int \frac{4 - y}{2} d y

\frac{1}{2} (4 y - \frac{y ^{2}}{2}) + C

Solution steps

\int \frac{4 - y}{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 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 4 d y - \int y d y)

\int 4 d y = 4 y

\int y d y = \frac{y ^{2}}{2}

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

Add a constant to the solution

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

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