What is the integral of (p/5-1/2) ?

The integral of (p/5-1/2) is (p^2)/(10)-1/2 p+C

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integral of (p/5-1/2)

\int (\frac{p}{5} - \frac{1}{2}) d p

\frac{p ^{2}}{10} - \frac{1}{2} p + C

Solution steps

\int \frac{p}{5} - \frac{1}{2} d p

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{p}{5} d p - \int \frac{1}{2} d p

\int \frac{p}{5} d p = \frac{p ^{2}}{10}

\int \frac{1}{2} d p = \frac{1}{2} p

= \frac{p ^{2}}{10} - \frac{1}{2} p

Add a constant to the solution

= \frac{p ^{2}}{10} - \frac{1}{2} p + C

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What is the integral of (p/5-1/2) ?
The integral of (p/5-1/2) is (p^2)/(10)-1/2 p+C