What is the integral from 1 to infinity of 6e^{-2x} ?

The integral from 1 to infinity of 6e^{-2x} is 3/(e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to infinity of 6e^{-2x}

\int_{1}^{\infty} 6 e^{- 2 x} d x

\frac{3}{e ^{2}}

Decimal

0.40600 \dots

Solution steps

\int_{1}^{\infty} 6 e^{- 2 x} d x

Compute the indefinite integral:

\int 6 e^{- 2 x} d x = - 3 e^{- 2 x} + C

Compute the boundaries:

\int_{1}^{\infty} 6 e^{- 2 x} d x = 0 - (- \frac{3}{e ^{2}})

= 0 - (- \frac{3}{e ^{2}})

Simplify

= \frac{3}{e ^{2}}

\int_{- 32}^{2} \frac{1}{∣ 2 x ∣} d x

y^{^{'}} = \frac{( y - x y )}{x ^{2}}

d er i v a t i v e ln (3 x)

\int 4 x ln (y) d x

t an g e n t - 3 x^{3} - 2 x^{2} - 1, at x = - 1

What is the integral from 1 to infinity of 6e^{-2x} ?
The integral from 1 to infinity of 6e^{-2x} is 3/(e^2)