What is the integral from 1 to infinity of 15e^{-5x} ?

The integral from 1 to infinity of 15e^{-5x} is 3/(e^5)

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 15e^{-5x}

\int_{1}^{\infty} 15 e^{- 5 x} d x

\frac{3}{e ^{5}}

Decimal Notation

0.02021 \dots

Solution steps

\int_{1}^{\infty} 15 e^{- 5 x} d x

Compute the indefinite integral:

\int 15 e^{- 5 x} d x = - 3 e^{- 5 x} + C

Compute the boundaries:

\int_{1}^{\infty} 15 e^{- 5 x} d x = 0 - (- \frac{3}{e ^{5}})

= 0 - (- \frac{3}{e ^{5}})

Simplify

= \frac{3}{e ^{5}}

\frac{d}{d x} (cos (2 x) + x^{2})

f (x) = (1 - x^{3}) + (1 - x^{3})

6 (x^{3} + 4)^{3}

\frac{d}{d x} (- 7 sin (\frac{x}{7}))

What is the integral from 1 to infinity of 15e^{-5x} ?
The integral from 1 to infinity of 15e^{-5x} is 3/(e^5)