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

The integral of x(sqrt(16-x^2)) is -1/3 (16-x^2)^{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(sqrt(16-x^2))

\int x (16 - x^{2}) d x

- \frac{1}{3} (16 - x^{2})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

= \int - \frac{u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int u d u

Apply the Power Rule

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

Substitute back

u = 16 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

x \to 2 lim (\frac{5 x}{x - 2})

\int e^{2 θ} sin (3 θ) d θ

(ln (f (x)))^{'}

10 s

(3^{- x^{2}})^{^{'}}

What is the integral of x(sqrt(16-x^2)) ?
The integral of x(sqrt(16-x^2)) is -1/3 (16-x^2)^{3/2}+C