What is the integral from 1 to infinity of xe^{-7x} ?

The integral from 1 to infinity of xe^{-7x} is 8/(49e^7)

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 xe^{-7x}

\int_{1}^{\infty} x e^{- 7 x} d x

\frac{8}{49 e ^{7}}

Decimal

0.00014 \dots

Solution steps

\int_{1}^{\infty} x e^{- 7 x} d x

Compute the indefinite integral:

\int x e^{- 7 x} d x = \frac{1}{49} (- 7 e^{- 7 x} x - e^{- 7 x}) + C

Compute the boundaries:

\int_{1}^{\infty} x e^{- 7 x} d x = 0 - (- \frac{8}{49 e ^{7}})

= 0 - (- \frac{8}{49 e ^{7}})

Simplify

= \frac{8}{49 e ^{7}}

y^{^{'}} + 4 y = 6 e^{- 3 t}

y \to - 3 lim (\frac{y ^{2} - 9}{2 y ^{2} + 7 y + 3})

16 y^{^{''}} - y = 0, y (- 7) = 1, y^{^{'}} (- 7) = - 1

\frac{d}{d x} (x^{\frac{5}{3}} - 5 x^{\frac{2}{3}})

\frac{d}{d x} (tan (x) + \frac{1}{x})

What is the integral from 1 to infinity of xe^{-7x} ?
The integral from 1 to infinity of xe^{-7x} is 8/(49e^7)