What is the integral of xsqrt(16-25x^2) ?

The integral of xsqrt(16-25x^2) is -1/75 (16-25x^2)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of xsqrt(16-25x^2)

\int x 16 - 25 x^{2} d x

- \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}} + C

Solution steps

\int x 16 - 25 x^{2} d x

Apply u-substitution

= \int - \frac{u}{50} d u

Take the constant out:

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

= - \frac{1}{50} \cdot \int u d u

Apply the Power Rule

= - \frac{1}{50} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 16 - 25 x^{2}

= - \frac{1}{50} \cdot \frac{2}{3} (16 - 25 x^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{50} \cdot \frac{2}{3} (16 - 25 x^{2})^{\frac{3}{2}} : - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}}

= - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{75} (16 - 25 x^{2})^{\frac{3}{2}} + C

What is the integral of xsqrt(16-25x^2) ?
The integral of xsqrt(16-25x^2) is -1/75 (16-25x^2)^{3/2}+C