What is the integral from 0 to 2 of e^{-2x+1} ?

The integral from 0 to 2 of e^{-2x+1} is (e^4-1)/(2e^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of e^{-2x+1}

\int_{0}^{2} e^{- 2 x + 1} d x

\frac{e ^{4} - 1}{2 e ^{3}}

Decimal

1.33424 \dots

Solution steps

\int_{0}^{2} e^{- 2 x + 1} d x

Apply u-substitution

: \int_{1}^{- 3} - \frac{1}{2} e^{u} d u

= \int_{1}^{- 3} - \frac{1}{2} e^{u} d u

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

= - \int_{- 3}^{1} - \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_{- 3}^{1} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{2} [e^{u}]_{- 3}^{1})

Simplify

= \frac{1}{2} [e^{u}]_{- 3}^{1}

Compute the boundaries:

e - \frac{1}{e ^{3}}

= \frac{1}{2} (e - \frac{1}{e ^{3}})

Simplify

= \frac{e ^{4} - 1}{2 e ^{3}}

\int_{1}^{2} π (e^{y})^{2} d y

\int_{- 2}^{3} e^{x + 2} d x

\int_{0}^{1} (x^{2} + 5 x) d x

\int_{0}^{2} e^{∣ 1 - x ∣} d x

\int_{0}^{1} \frac{1}{x ^{2} + x + 1} d x

What is the integral from 0 to 2 of e^{-2x+1} ?
The integral from 0 to 2 of e^{-2x+1} is (e^4-1)/(2e^3)