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

The integral from 0 to infinity of e^{-2y} is 1/2

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 e^{-2y}

\int_{0}^{\infty} e^{- 2 y} d y

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{\infty} e^{- 2 y} d y

Compute the indefinite integral:

\int e^{- 2 y} d y = - \frac{1}{2} e^{- 2 y} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- 2 y} d y = 0 - (- \frac{1}{2})

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

\int_{4}^{9} x d x

\int_{- 2}^{1} - x^{2} + 5 d x

\int_{x - z}^{x + z} z d z

\int_{0}^{\frac{π}{2}} 3 cos (x) sin (x) d x

\int_{0}^{3} ln (x) d x

What is the integral from 0 to infinity of e^{-2y} ?
The integral from 0 to infinity of e^{-2y} is 1/2