What is the integral from 0 to 2 of (6-x-y)/8 ?

The integral from 0 to 2 of (6-x-y)/8 is (-y+5)/4

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integral from 0 to 2 of (6-x-y)/8

\int_{0}^{2} \frac{6 - x - y}{8} d x

\frac{- y + 5}{4}

Solution steps

\int_{0}^{2} \frac{6 - x - y}{8} d x

Take the constant out:

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

= \frac{1}{8} \cdot \int_{0}^{2} 6 - y - x d x

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}{8} (\int_{0}^{2} 6 d x - \int_{0}^{2} y d x - \int_{0}^{2} x d x)

\int_{0}^{2} 6 d x = 12

\int_{0}^{2} y d x = 2 y

\int_{0}^{2} x d x = 2

= \frac{1}{8} (12 - 2 y - 2)

Simplify

\frac{1}{8} (12 - 2 y - 2) : \frac{- y + 5}{4}

= \frac{- y + 5}{4}

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What is the integral from 0 to 2 of (6-x-y)/8 ?
The integral from 0 to 2 of (6-x-y)/8 is (-y+5)/4