What is the integral from 6 to infinity of xe^{-2x} ?

The integral from 6 to infinity of xe^{-2x} is (13)/(4e^{12)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 6 to infinity of xe^{-2x}

\int_{6}^{\infty} x e^{- 2 x} d x

\frac{13}{4 e ^{12}}

Decimal Notation

0.00001 \dots

Solution steps

\int_{6}^{\infty} x e^{- 2 x} d x

Compute the indefinite integral:

\int x e^{- 2 x} d x = \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

Compute the boundaries:

\int_{6}^{\infty} x e^{- 2 x} d x = 0 - (- \frac{13}{4 e ^{12}})

= 0 - (- \frac{13}{4 e ^{12}})

Simplify

= \frac{13}{4 e ^{12}}

\int - 5 cos (x) d x

y^{^{''}} - 8 y^{^{'}} + 10 y = 0

\int 8 sec^{2} (x) d x

x \to 3 lim (f (x) (g (x)))

d y = (y - 1)^{2} d x

What is the integral from 6 to infinity of xe^{-2x} ?
The integral from 6 to infinity of xe^{-2x} is (13)/(4e^{12)}