What is the integral from 0 to 8 of 1/(2x^{1/3)} ?

The integral from 0 to 8 of 1/(2x^{1/3)} is 3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 8 of 1/(2x^{1/3)}

\int_{0}^{8} \frac{1}{2 x ^{\frac{1}{3}}} d x

3

Solution steps

\int_{0}^{8} \frac{1}{2 x ^{\frac{1}{3}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{8} \frac{1}{x ^{\frac{1}{3}}} d x

Apply the Power Rule

: [\frac{3}{2} x^{\frac{2}{3}}]_{0}^{8}

= \frac{1}{2} [\frac{3}{2} x^{\frac{2}{3}}]_{0}^{8}

Compute the boundaries:

6

= \frac{1}{2} \cdot 6

Simplify

= 3

\int_{0}^{2} (- x^{4} + 2 x^{2} + 8) d x

\int_{0}^{5} 3 x - 8 d x

\int_{0}^{1} t e^{- t^{2}} d t

\int_{0}^{2} 13 x (x^{2} + 1)^{2} d x

\int_{0}^{1} (1 - x) cos (nπ x) d x

What is the integral from 0 to 8 of 1/(2x^{1/3)} ?
The integral from 0 to 8 of 1/(2x^{1/3)} is 3