What is the integral from 0 to infinity of 5e^{-2x} ?

The integral from 0 to infinity of 5e^{-2x} is 5/2

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^{-2x}

\int_{0}^{\infty} 5 e^{- 2 x} d x

\frac{5}{2}

Decimal

2.5

Solution steps

\int_{0}^{\infty} 5 e^{- 2 x} d x

Compute the indefinite integral:

\int 5 e^{- 2 x} d x = - \frac{5}{2} e^{- 2 x} + C

Compute the boundaries:

\int_{0}^{\infty} 5 e^{- 2 x} d x = 0 - (- \frac{5}{2})

= 0 - (- \frac{5}{2})

Simplify

= \frac{5}{2}

\int e^{- 2 x} tan (e^{- 2 x}) d x

y^{^{''}} - y^{^{'}} + y = cos (2 t) + sin (2 t)

\frac{1}{s ( s ^{2} + 6 π s + π ^{2} )}

\int \frac{3 tan ( x ) - 4 cos ^{2} ( x )}{cos ( x )} d x

5 x^{2} (3 x^{2} - 1)^{2}

What is the integral from 0 to infinity of 5e^{-2x} ?
The integral from 0 to infinity of 5e^{-2x} is 5/2