What is the integral of (20-x)/4 ?

The integral of (20-x)/4 is 1/4 (20x-(x^2)/2)+C

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integral of (20-x)/4

\int \frac{20 - x}{4} d x

\frac{1}{4} (20 x - \frac{x ^{2}}{2}) + C

Solution steps

\int \frac{20 - x}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int 20 - 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}{4} (\int 20 d x - \int x d x)

\int 20 d x = 20 x

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

= \frac{1}{4} (20 x - \frac{x ^{2}}{2})

Add a constant to the solution

= \frac{1}{4} (20 x - \frac{x ^{2}}{2}) + C

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What is the integral of (20-x)/4 ?
The integral of (20-x)/4 is 1/4 (20x-(x^2)/2)+C