What is the integral of (4x^2-3x) ?

The integral of (4x^2-3x) is (4x^3)/3-(3x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (4x^2-3x)

\int (4 x^{2} - 3 x) d x

\frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} + C

Solution steps

\int 4 x^{2} - 3 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 4 x^{2} d x - \int 3 x d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

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

= \frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2}

Add a constant to the solution

= \frac{4 x ^{3}}{3} - \frac{3 x ^{2}}{2} + C

What is the integral of (4x^2-3x) ?
The integral of (4x^2-3x) is (4x^3)/3-(3x^2)/2+C