What is the integral of x^2sqrt(64-x^3) ?

The integral of x^2sqrt(64-x^3) is -2/9 (64-x^3)^{3/2}+C

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integral of x^2sqrt(64-x^3)

\int x^{2} 64 - x^{3} d x

- \frac{2}{9} (64 - x^{3})^{\frac{3}{2}} + C

Solution steps

\int x^{2} 64 - x^{3} d x

Apply u-substitution

: \int - \frac{u}{3} d u

= \int - \frac{u}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 64 - x^{3}

= - \frac{1}{3} \cdot \frac{2}{3} (64 - x^{3})^{\frac{3}{2}}

Simplify

- \frac{1}{3} \cdot \frac{2}{3} (64 - x^{3})^{\frac{3}{2}} : - \frac{2}{9} (64 - x^{3})^{\frac{3}{2}}

= - \frac{2}{9} (64 - x^{3})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{9} (64 - x^{3})^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(64-x^3) ?
The integral of x^2sqrt(64-x^3) is -2/9 (64-x^3)^{3/2}+C