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

- 0.63212 \dots

Solution steps

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

Take the constant out:

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

= - \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})

Simplify

= - 1 + \frac{1}{e}

\frac{d y}{d x} = 2 x + 3

x \to 0 - lim (\frac{2}{3 x ^{\frac{1}{3}}})

\int tan (4 θ) d θ

f (x) = 4 e^{x}

x \to - \infty lim (∣ x ∣ - 3)

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