What is the integral from 1 to infinity of x^{-1/3} ?

The integral from 1 to infinity of x^{-1/3} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of x^{-1/3}

\int_{1}^{\infty} x^{- \frac{1}{3}} d x

diverges

Solution steps

\int_{1}^{\infty} x^{- \frac{1}{3}} d x

Compute the indefinite integral:

\int x^{- \frac{1}{3}} d x = \frac{3}{2} x^{\frac{2}{3}} + C

Compute the boundaries:

\int_{1}^{\infty} x^{- \frac{1}{3}} d x =

diverges

= diverges

\frac{d}{d t} (- e^{- 4 t} + 4 t e^{- 4 t})

\frac{( z + 2 )}{z ^{2}}

\int \frac{x}{x ^{2} + 6 x + 8} d x

\frac{d}{d x} (\frac{x ^{2} + x - 1}{x ^{2} - 1})

\int 6 cot^{4} (x) d x

What is the integral from 1 to infinity of x^{-1/3} ?
The integral from 1 to infinity of x^{-1/3} is diverges