What is the integral from 0 to b/M of (v^2M^2-b) ?

The integral from 0 to b/M of (v^2M^2-b) is (b^3}{3M}-\frac{b^2)/M

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integral from 0 to b/M of (v^2M^2-b)

\int_{0}^{\frac{b}{M}} (v^{2} M^{2} - b) d v

\frac{b ^{3}}{3 M} - \frac{b ^{2}}{M}

Solution steps

\int_{0}^{\frac{b}{M}} v^{2} M^{2} - b d v

Apply the Sum Rule:

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

= \int_{0}^{\frac{b}{M}} v^{2} M^{2} d v - \int_{0}^{\frac{b}{M}} b d v

\int_{0}^{\frac{b}{M}} v^{2} M^{2} d v = \frac{b ^{3}}{3 M}

\int_{0}^{\frac{b}{M}} b d v = \frac{b ^{2}}{M}

= \frac{b ^{3}}{3 M} - \frac{b ^{2}}{M}

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What is the integral from 0 to b/M of (v^2M^2-b) ?
The integral from 0 to b/M of (v^2M^2-b) is (b^3}{3M}-\frac{b^2)/M