What is the integral from-1 to 1 of 1/(e^x) ?

The integral from-1 to 1 of 1/(e^x) is -1/e+e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 1/(e^x)

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

- \frac{1}{e} + e

Decimal

2.35040 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [- \frac{1}{u}]_{\frac{1}{e}}^{e}

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

Compute the boundaries:

- \frac{1}{e} + e

= - \frac{1}{e} + e

\int_{0}^{7} (x) d x

\int_{5}^{x} \frac{2}{5} - \frac{1}{25} x d x

\int_{3}^{8} x d x

\int_{0}^{1} (\frac{4}{1 + x ^{2}}) d x

\int_{0}^{π} x (sin (x)) d x

What is the integral from-1 to 1 of 1/(e^x) ?
The integral from-1 to 1 of 1/(e^x) is -1/e+e