What is the integral of (z^2)/(1^3)+z^3 ?

The integral of (z^2)/(1^3)+z^3 is (z^3)/3+(z^4)/4+C

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integral of (z^2)/(1^3)+z^3

\int \frac{z ^{2}}{1 ^{3}} + z^{3}

\frac{z ^{3}}{3} + \frac{z ^{4}}{4} + C

Solution steps

\int \frac{z ^{2}}{1 ^{3}} + z^{3} d z

Simplify

= \int z^{2} + z^{3} d z

Apply the Sum Rule:

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

= \int z^{2} d z + \int z^{3} d z

\int z^{2} d z = \frac{z ^{3}}{3}

\int z^{3} d z = \frac{z ^{4}}{4}

= \frac{z ^{3}}{3} + \frac{z ^{4}}{4}

Add a constant to the solution

= \frac{z ^{3}}{3} + \frac{z ^{4}}{4} + C

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f (r) = \frac{4}{3} π r^{3}

What is the integral of (z^2)/(1^3)+z^3 ?
The integral of (z^2)/(1^3)+z^3 is (z^3)/3+(z^4)/4+C