What is the integral from 0 to 3 of (r^2+z)r ?

The integral from 0 to 3 of (r^2+z)r is (9z)/2+81/4

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integral from 0 to 3 of (r^2+z)r

\int_{0}^{3} (r^{2} + z) r d r

\frac{9 z}{2} + \frac{81}{4}

Solution steps

\int_{0}^{3} (r^{2} + z) r d r

Expand

(z + r^{2}) r : zr + r^{3}

= \int_{0}^{3} zr + r^{3} 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}^{3} zr d r + \int_{0}^{3} r^{3} d r

\int_{0}^{3} zr d r = \frac{9 z}{2}

\int_{0}^{3} r^{3} d r = \frac{81}{4}

= \frac{9 z}{2} + \frac{81}{4}

What is the integral from 0 to 3 of (r^2+z)r ?
The integral from 0 to 3 of (r^2+z)r is (9z)/2+81/4