What is the integral of 8sqrt(36-4x)-32 ?

The integral of 8sqrt(36-4x)-32 is -32/3 (9-x)^{3/2}-32x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8sqrt(36-4x)-32

\int 8 36 - 4 x - 32 d x

- \frac{32}{3} (9 - x)^{\frac{3}{2}} - 32 x + C

Solution steps

\int 8 36 - 4 x - 32 d x

8 36 - 4 x = 16 9 - x

= \int 16 9 - x - 32 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 16 9 - x d x - \int 32 d x

\int 16 9 - x d x = - \frac{32}{3} (9 - x)^{\frac{3}{2}}

\int 32 d x = 32 x

= - \frac{32}{3} (9 - x)^{\frac{3}{2}} - 32 x

Add a constant to the solution

= - \frac{32}{3} (9 - x)^{\frac{3}{2}} - 32 x + C

\int \frac{4 + 5 x}{1 + x ^{2}} d x

\frac{d}{d t} ((90 cos (\frac{π}{6})) t)

\int \frac{ln ( 21 x )}{x} d x

\frac{d}{d x} (\frac{x}{9 x + 2})

\int (9 - 5 x) d x

What is the integral of 8sqrt(36-4x)-32 ?
The integral of 8sqrt(36-4x)-32 is -32/3 (9-x)^{3/2}-32x+C