What is the integral from 0 to ln(6) of e^{x-e^x} ?

The integral from 0 to ln(6) of e^{x-e^x} is 1/e-1/(e^6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to ln(6) of e^{x-e^x}

\int_{0}^{l n (6)} e^{x - e^{x}} d x

\frac{1}{e} - \frac{1}{e ^{6}}

Decimal

0.36540 \dots

Solution steps

\int_{0}^{l n (6)} e^{x - e^{x}} d x

Apply u-substitution

: \int_{1}^{6} e^{- u} d u

= \int_{1}^{6} e^{- u} d u

Apply u-substitution

: \int_{- 1}^{- 6} - e^{v} d v

= \int_{- 1}^{- 6} - e^{v} d v

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

= - \int_{- 6}^{- 1} - e^{v} d v

Take the constant out:

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

= - (- \int_{- 6}^{- 1} e^{v} d v)

Use the common integral:

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

= - (- [e^{v}]_{- 6}^{- 1})

Simplify

= [e^{v}]_{- 6}^{- 1}

Compute the boundaries:

\frac{1}{e} - \frac{1}{e ^{6}}

= \frac{1}{e} - \frac{1}{e ^{6}}

\int_{- 2}^{2} (x + ∣ x ∣) e^{- ∣ x ∣} d x

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

\int_{- π}^{0} 0 cos (n x) d x

\int_{9}^{16} (x - 2) d x

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

What is the integral from 0 to ln(6) of e^{x-e^x} ?
The integral from 0 to ln(6) of e^{x-e^x} is 1/e-1/(e^6)