What is the integral from-infinity to 1 of-e^x ?

The integral from-infinity to 1 of-e^x is -e

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-infinity to 1 of-e^x

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

- e

Decimal Notation

- 2.71828 \dots

Solution steps

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

Compute the indefinite integral:

\int - e^{x} d x = - e^{x} + C

Compute the boundaries:

\int_{- \infty}^{1} - e^{x} d x = - e - 0

= - e - 0

Simplify

= - e

\int (\frac{r}{3}) cos (\frac{r}{3}) d r

\frac{6 ( s + 1 )}{s ^{2} ( s + 1 )}

\frac{\partial}{\partial x} (arcsin (2 x))

\int (x y) d x

y^{^{''}} + 81 y = 0, y (0) = 3, y^{^{'}} (0) = 5

What is the integral from-infinity to 1 of-e^x ?
The integral from-infinity to 1 of-e^x is -e