What is the integral from 0 to infinity of 5e^{-3x} ?

The integral from 0 to infinity of 5e^{-3x} is 5/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of 5e^{-3x}

\int_{0}^{\infty} 5 e^{- 3 x} d x

\frac{5}{3}

Decimal Notation

1.66666 \dots

Solution steps

\int_{0}^{\infty} 5 e^{- 3 x} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

= \frac{5}{3}

\int_{\frac{1}{2} x}^{1} x^{2} e^{x y} d y

\int_{4}^{5} \frac{2}{( x - 3 )} d x

\int_{0}^{12} 9800 \cdot 9 π \cdot (12 - x) d x

\int_{- 1}^{2} (x + x^{4}) d x

\int_{0}^{2} 0.5 e^{- 0.5 x} d x

What is the integral from 0 to infinity of 5e^{-3x} ?
The integral from 0 to infinity of 5e^{-3x} is 5/3