What is the integral from 0 to y of (r^2-(r^3)/x) ?

The integral from 0 to y of (r^2-(r^3)/x) is -(y^4}{4x}+\frac{y^3)/3

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integral from 0 to y of (r^2-(r^3)/x)

\int_{0}^{y} (r^{2} - \frac{r ^{3}}{x}) d r

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

Solution steps

\int_{0}^{y} r^{2} - \frac{r ^{3}}{x} d r

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}^{y} \frac{r ^{3}}{x} d r + \int_{0}^{y} r^{2} d r

\int_{0}^{y} \frac{r ^{3}}{x} d r = \frac{y ^{4}}{4 x}

\int_{0}^{y} r^{2} d r = \frac{y ^{3}}{3}

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

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What is the integral from 0 to y of (r^2-(r^3)/x) ?
The integral from 0 to y of (r^2-(r^3)/x) is -(y^4}{4x}+\frac{y^3)/3