What is the integral from 2 to infinity of ve^{-3v} ?

The integral from 2 to infinity of ve^{-3v} is 7/(9e^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to infinity of ve^{-3v}

\int_{2}^{\infty} v e^{- 3 v} d v

\frac{7}{9 e ^{6}}

Decimal

0.00192 \dots

Solution steps

\int_{2}^{\infty} v e^{- 3 v} d v

Compute the indefinite integral:

\int v e^{- 3 v} d v = \frac{1}{9} (- 3 e^{- 3 v} v - e^{- 3 v}) + C

Compute the boundaries:

\int_{2}^{\infty} v e^{- 3 v} d v = 0 - (- \frac{7}{9 e ^{6}})

= 0 - (- \frac{7}{9 e ^{6}})

Simplify

= \frac{7}{9 e ^{6}}

\int_{- 1}^{1} (1 + 6 x y^{2}) d y

\int_{0}^{4} - 6 x^{2} - 12 x + 144 d x

\int_{0}^{b} 7 e^{x} d x

\int_{0}^{x} cos (9 t) d t

\int_{0}^{6} x^{2} - 3 x d x

What is the integral from 2 to infinity of ve^{-3v} ?
The integral from 2 to infinity of ve^{-3v} is 7/(9e^6)