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

The integral from 0 to 4 of (x^2-sqrt(x)) is 16

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-sqrt(x))

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

16

Solution steps

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

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

\int_{0}^{4} x d x = \frac{16}{3}

= \frac{64}{3} - \frac{16}{3}

Simplify

\frac{64}{3} - \frac{16}{3} : 16

= 16

What is the integral from 0 to 4 of (x^2-sqrt(x)) ?
The integral from 0 to 4 of (x^2-sqrt(x)) is 16