What is the integral of 9-(x^2)/3 ?

The integral of 9-(x^2)/3 is 9x-(x^3)/9+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 9-(x^2)/3

\int 9 - \frac{x ^{2}}{3} d x

9 x - \frac{x ^{3}}{9} + C

Solution steps

\int 9 - \frac{x ^{2}}{3} 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 9 d x - \int \frac{x ^{2}}{3} d x

\int 9 d x = 9 x

\int \frac{x ^{2}}{3} d x = \frac{x ^{3}}{9}

= 9 x - \frac{x ^{3}}{9}

Add a constant to the solution

= 9 x - \frac{x ^{3}}{9} + C