What is the integral of (-(ARx^2)/2+C) ?

The integral of (-(ARx^2)/2+C) is -(ARx^3)/6+Cx+C

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integral of (-(ARx^2)/2+C)

\int (- \frac{A R x ^{2}}{2} + C) d x

- \frac{A R x ^{3}}{6} + C x + C

Solution steps

\int - \frac{A R x ^{2}}{2} + C 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{A R x ^{2}}{2} d x + \int C d x

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

\int C d x = C x

= - \frac{A R x ^{3}}{6} + C x

Add a constant to the solution

= - \frac{A R x ^{3}}{6} + C x + C

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What is the integral of (-(ARx^2)/2+C) ?
The integral of (-(ARx^2)/2+C) is -(ARx^3)/6+Cx+C