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

The integral from 0 to 1 of e^{-sqrt(x)} is -2(-1+2/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^{-sqrt(x)}

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

- 2 (- 1 + \frac{2}{e})

Decimal Notation

0.52848 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 1} 2 e^{u} u d u

= \int_{0}^{- 1} 2 e^{u} u d u

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

= - \int_{- 1}^{0} 2 e^{u} u d u

Take the constant out:

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

= - 2 \cdot \int_{- 1}^{0} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 1}^{0}

= - 2 [e^{u} u - \int e^{u} d u]_{- 1}^{0}

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

= - 2 [e^{u} u - e^{u}]_{- 1}^{0}

Compute the boundaries:

- 1 + \frac{2}{e}

= - 2 (- 1 + \frac{2}{e})

\int_{0}^{4} x + 2 d x

\int_{2}^{4} (- x^{2} + 2 x + 8) d x

\int_{0}^{1} (x^{12} + 1 2^{x}) d x

\int_{0}^{1} x^{4} - x^{5} d x

\int_{1}^{3} x^{2} + 3 d x

What is the integral from 0 to 1 of e^{-sqrt(x)} ?
The integral from 0 to 1 of e^{-sqrt(x)} is -2(-1+2/e)