What is the integral from 1 to infinity of 9e^{-3x} ?

The integral from 1 to infinity of 9e^{-3x} is 3/(e^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of 9e^{-3x}

\int_{1}^{\infty} 9 e^{- 3 x} d x

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

Decimal Notation

0.14936 \dots

Solution steps

\int_{1}^{\infty} 9 e^{- 3 x} d x

Compute the indefinite integral:

\int 9 e^{- 3 x} d x = - 3 e^{- 3 x} + C

Compute the boundaries:

\int_{1}^{\infty} 9 e^{- 3 x} d x = 0 - (- \frac{3}{e ^{3}})

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

Simplify

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

y = 5 (x^{2} + 2 x)^{3}

\frac{d}{d x} (- ln (2) \cdot 2^{- x})

\int_{0}^{3} x x + 1 d x

f (x) = 9 x - 8 x, at x = 1

\frac{d}{d t} (e^{4 (e^{t} - 1)})

What is the integral from 1 to infinity of 9e^{-3x} ?
The integral from 1 to infinity of 9e^{-3x} is 3/(e^3)