What is the integral from 1 to x of (5-x)/8 ?

The integral from 1 to x of (5-x)/8 is 1/16 (-x^2+10x-9)

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integral from 1 to x of (5-x)/8

\int_{1}^{x} \frac{5 - x}{8} d x

\frac{1}{16} (- x^{2} + 10 x - 9)

Solution steps

\int_{1}^{x} \frac{5 - x}{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_{1}^{x} 5 - 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_{1}^{x} 5 d x - \int_{1}^{x} x d x)

\int_{1}^{x} 5 d x = 5 x - 5

\int_{1}^{x} x d x = \frac{x ^{2} - 1}{2}

= \frac{1}{8} (5 x - 5 - \frac{x ^{2} - 1}{2})

Simplify

\frac{1}{8} (5 x - 5 - \frac{x ^{2} - 1}{2}) : \frac{1}{16} (- x^{2} + 10 x - 9)

= \frac{1}{16} (- x^{2} + 10 x - 9)

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What is the integral from 1 to x of (5-x)/8 ?
The integral from 1 to x of (5-x)/8 is 1/16 (-x^2+10x-9)