What is the integral from 2 to 3 of e^{1-x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 2 to 3 of e^{1-x}

\int_{2}^{3} e^{1 - x} d x

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

Decimal

0.23254 \dots

Solution steps

\int_{2}^{3} e^{1 - x} d x

Apply u-substitution

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

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