What is the integral from 1 to infinity of ne^{-n^2} ?

The integral from 1 to infinity of ne^{-n^2} 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 ne^{-n^2}

\int_{1}^{\infty} n e^{- n^{2}}

diverges

Solution steps

\int_{1}^{\infty} n e^{- n^{2}} d x

Compute the indefinite integral:

\int n e^{- n^{2}} d x = n e^{- n^{2}} x + C

Compute the boundaries:

\int_{1}^{\infty} n e^{- n^{2}} d x =

diverges

= diverges

\int_{1}^{4} 5 x (x - 1) d x

\int_{0}^{1} y \cdot 2 y d y

\int_{1}^{2} \frac{1}{t - 1} d t

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

\int_{0}^{π} (8 + 3 sin (x)) d x

What is the integral from 1 to infinity of ne^{-n^2} ?
The integral from 1 to infinity of ne^{-n^2} is diverges