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

The integral from 1 to infinity of x^{-p} is Undefined-1/(-p+1)

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^{-p}

\int_{1}^{\infty} x^{- p} d x

Undefined - \frac{1}{- p + 1}

Solution steps

\int_{1}^{\infty} x^{- p} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} x^{- p} d x =

Undefined

- \frac{1}{- p + 1}

= Undefined - \frac{1}{- p + 1}

\int_{- 5}^{6} ∣ x ∣ d x

\int_{- \infty}^{\infty} e^{x^{2}} d x

\int_{0}^{3} \frac{1}{( x - 2 ) ^{3}} d x

\int_{0}^{\frac{π}{4}} sec (θ) tan (θ) d θ

\int_{0}^{1} 2 e^{2 x} d x

What is the integral from 1 to infinity of x^{-p} ?
The integral from 1 to infinity of x^{-p} is Undefined-1/(-p+1)