What is the integral of-(2sqrt(z))/3 ?

The integral of-(2sqrt(z))/3 is -4/9 z^{3/2}+C

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integral of-(2sqrt(z))/3

\int - \frac{2 z}{3} d z

- \frac{4}{9} z^{\frac{3}{2}} + C

Solution steps

\int - \frac{2 z}{3} d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{2}{3} \cdot \int z d z

Apply the Power Rule

: \frac{2}{3} z^{\frac{3}{2}}

= - \frac{2}{3} \cdot \frac{2}{3} z^{\frac{3}{2}}

Simplify

- \frac{2}{3} \cdot \frac{2}{3} z^{\frac{3}{2}} : - \frac{4}{9} z^{\frac{3}{2}}

= - \frac{4}{9} z^{\frac{3}{2}}

Add a constant to the solution

= - \frac{4}{9} z^{\frac{3}{2}} + C

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What is the integral of-(2sqrt(z))/3 ?
The integral of-(2sqrt(z))/3 is -4/9 z^{3/2}+C