What is the integral from 8 to infinity of xe^{-2x} ?

The integral from 8 to infinity of xe^{-2x} is (17)/(4e^{16)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 8 to infinity of xe^{-2x}

\int_{8}^{\infty} x e^{- 2 x} d x

\frac{17}{4 e ^{16}}

Decimal Notation

4.78274 E - 7

Solution steps

\int_{8}^{\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_{8}^{\infty} x e^{- 2 x} d x = 0 - (- \frac{17}{4 e ^{16}})

= 0 - (- \frac{17}{4 e ^{16}})

Simplify

= \frac{17}{4 e ^{16}}

\frac{d}{d x} (sin^{3} (x) + sin (3 x))

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

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

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

\frac{1}{( s - 4 )}

What is the integral from 8 to infinity of xe^{-2x} ?
The integral from 8 to infinity of xe^{-2x} is (17)/(4e^{16)}