What is the integral from 0 to 4 of (x^2-4x) ?

The integral from 0 to 4 of (x^2-4x) is -32/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 4 of (x^2-4x)

\int_{0}^{4} (x^{2} - 4 x) d x

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

\int_{0}^{4} x^{2} - 4 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_{0}^{4} x^{2} d x - \int_{0}^{4} 4 x d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 4 x d x = 32

= \frac{64}{3} - 32

Simplify

\frac{64}{3} - 32 : - \frac{32}{3}

= - \frac{32}{3}

What is the integral from 0 to 4 of (x^2-4x) ?
The integral from 0 to 4 of (x^2-4x) is -32/3