What is the integral from 1 to 2 of e^{-2x} ?

The integral from 1 to 2 of e^{-2x} is (e^2-1)/(2e^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of e^{-2x}

\int_{1}^{2} e^{- 2 x} d x

\frac{e ^{2} - 1}{2 e ^{4}}

Decimal Notation

0.05850 \dots

Solution steps

\int_{1}^{2} e^{- 2 x} d x

Apply u-substitution

: \int_{- 2}^{- 4} - \frac{1}{2} e^{u} d u

= \int_{- 2}^{- 4} - \frac{1}{2} e^{u} d u

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

= - \int_{- 4}^{- 2} - \frac{1}{2} e^{u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{- 4}^{- 2} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 4}^{- 2})

Simplify

= \frac{1}{2} [e^{u}]_{- 4}^{- 2}

Compute the boundaries:

\frac{1}{e ^{2}} - \frac{1}{e ^{4}}

= \frac{1}{2} (\frac{1}{e ^{2}} - \frac{1}{e ^{4}})

Simplify

= \frac{e ^{2} - 1}{2 e ^{4}}

\int_{1}^{3} 5 x d x

\int_{0}^{\infty} t e^{- pt} d t

\int_{0}^{4} (x - 3) (x - 1) d x

\int_{0}^{1} z e^{- z^{2}} d z

\int_{- 1}^{3} \frac{1}{x ^{3}} d x

What is the integral from 1 to 2 of e^{-2x} ?
The integral from 1 to 2 of e^{-2x} is (e^2-1)/(2e^4)