What is the integral of (1/3)x^3-5x ?

The integral of (1/3)x^3-5x is (x^4}{12}-\frac{5x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1/3)x^3-5x

\int (\frac{1}{3}) x^{3} - 5 x d x

\frac{x ^{4}}{12} - \frac{5 x ^{2}}{2} + C

Solution steps

\int \frac{1}{3} x^{3} - 5 x 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 \frac{1}{3} x^{3} d x - \int 5 x d x

\int \frac{1}{3} x^{3} d x = \frac{x ^{4}}{12}

\int 5 x d x = \frac{5 x ^{2}}{2}

= \frac{x ^{4}}{12} - \frac{5 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{12} - \frac{5 x ^{2}}{2} + C

What is the integral of (1/3)x^3-5x ?
The integral of (1/3)x^3-5x is (x^4}{12}-\frac{5x^2)/2+C