What is the integral from 0 to 1 of e^{-x} ?

The integral from 0 to 1 of e^{-x} is 1-1/e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of e^{-x}

\int_{0}^{1} e^{- x} d x

1 - \frac{1}{e}

Decimal Notation

0.63212 \dots

Solution steps

\int_{0}^{1} e^{- x} d x

Apply u-substitution

: \int_{0}^{- 1} - e^{u} d u

= \int_{0}^{- 1} - e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 1}^{0} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- [e^{u}]_{- 1}^{0})

Simplify

= [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

\int 2 x x^{2} + 3 d x

\int 2 sin (d) x d x

\int \frac{1}{( 2 x - 1 ) ( x + 3 )} d x

\frac{\partial}{\partial x} (3 sin (3 x))

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