What is the integral of x^2sqrt(16-x^3) ?

The integral of x^2sqrt(16-x^3) is -2/9 (16-x^3)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2sqrt(16-x^3)

\int x^{2} 16 - x^{3} d x

- \frac{2}{9} (16 - x^{3})^{\frac{3}{2}} + C

Solution steps

\int x^{2} 16 - 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 = 16 - x^{3}

= - \frac{1}{3} \cdot \frac{2}{3} (16 - x^{3})^{\frac{3}{2}}

Simplify

- \frac{1}{3} \cdot \frac{2}{3} (16 - x^{3})^{\frac{3}{2}} : - \frac{2}{9} (16 - x^{3})^{\frac{3}{2}}

= - \frac{2}{9} (16 - x^{3})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{9} (16 - x^{3})^{\frac{3}{2}} + C

\int e^{2 x} sin (\frac{x}{2}) d x

\int (x + 2) (x^{2} - 1) d x

\int 11 x^{- 6} d x

\int 4 x (4 x + 1)^{3} d x

\int (3 x^{2} + 2 x) e^{x^{3} + x^{2}} d x

What is the integral of x^2sqrt(16-x^3) ?
The integral of x^2sqrt(16-x^3) is -2/9 (16-x^3)^{3/2}+C