What is the integral from 1 to infinity of 3xe^{-x} ?

The integral from 1 to infinity of 3xe^{-x} is 6/e

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integral from 1 to infinity of 3xe^{-x}

\int_{1}^{\infty} 3 x e^{- x} d x

\frac{6}{e}

Decimal Notation

2.20727 \dots

Solution steps

\int_{1}^{\infty} 3 x e^{- x} d x

Compute the indefinite integral:

\int 3 x e^{- x} d x = 3 (- e^{- x} x - e^{- x}) + C

Compute the boundaries:

\int_{1}^{\infty} 3 x e^{- x} d x = 0 - (- \frac{6}{e})

= 0 - (- \frac{6}{e})

Simplify

= \frac{6}{e}

\int tanh^{2} (3 x) d x

n = 0 \sum \infty n \cdot a^{n}

\frac{d}{d z} (e^{x z})

e^{- x + 1} x - 2 e^{- x + 1}

f (x) = \frac{5}{x}, (- 4, - \frac{5}{4})

What is the integral from 1 to infinity of 3xe^{-x} ?
The integral from 1 to infinity of 3xe^{-x} is 6/e