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

The integral from 2 to 4 of (6-x-y)/8 is (-x+3)/4

integral from 2 to 4 of (6-x-y)/8

\int_{2}^{4} \frac{6 - x - y}{8} d y

\frac{- x + 3}{4}

Solution steps

\int_{2}^{4} \frac{6 - x - y}{8} d y

Take the constant out:

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

= \frac{1}{8} \cdot \int_{2}^{4} 6 - x - 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}{8} (\int_{2}^{4} 6 d y - \int_{2}^{4} x d y - \int_{2}^{4} y d y)

\int_{2}^{4} 6 d y = 12

\int_{2}^{4} x d y = 2 x

\int_{2}^{4} y d y = 6

= \frac{1}{8} (12 - 2 x - 6)

Simplify

\frac{1}{8} (12 - 2 x - 6) : \frac{- x + 3}{4}

= \frac{- x + 3}{4}

in v erse l a pl a ce \frac{2}{s + 1} + \frac{8 s}{s ^{2} + 9}

x \to 0 lim (\frac{e}{x})

\int \frac{1}{( x + y )} d x

d er i v a t i v e (x + 1)^{3}

in v erse l a pl a ce t

What is the integral from 2 to 4 of (6-x-y)/8 ?
The integral from 2 to 4 of (6-x-y)/8 is (-x+3)/4