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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{e ^{\frac{5}{2}}}

Decimal

0.16416 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

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

Simplify

= \frac{2}{e ^{\frac{5}{2}}}

\frac{d}{d x} (\frac{2}{3} x^{3} - 4 x^{2} + 10 x + 6)

\frac{d y}{d x} = \frac{y}{x}, y (1) = 2

\int_{0}^{10} 200 - 4 x d x

y^{^{'}} = 8 x y, y (1) = 1

f (x) = - 0.004 x^{3} - 0.3 x^{2} + 600 x - 800

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