What is the integral of (((-x^2+8)/2)-((x^2)/2)) ?

The integral of (((-x^2+8)/2)-((x^2)/2)) is -(x^3)/3+4x+C

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integral of (((-x^2+8)/2)-((x^2)/2))

\int ((\frac{- x ^{2} + 8}{2}) - (\frac{x ^{2}}{2})) d x

- \frac{x ^{3}}{3} + 4 x + C

Solution steps

\int \frac{- x ^{2} + 8}{2} - \frac{x ^{2}}{2} 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

= \int \frac{- x ^{2} + 8}{2} d x - \int \frac{x ^{2}}{2} d x

\int \frac{- x ^{2} + 8}{2} d x = \frac{1}{2} (- \frac{x ^{3}}{3} + 8 x)

\int \frac{x ^{2}}{2} d x = \frac{x ^{3}}{6}

= \frac{1}{2} (- \frac{x ^{3}}{3} + 8 x) - \frac{x ^{3}}{6}

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (((-x^2+8)/2)-((x^2)/2)) ?
The integral of (((-x^2+8)/2)-((x^2)/2)) is -(x^3)/3+4x+C