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

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

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^{2x}

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

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

Decimal Notation

0.06766 \dots

Solution steps

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

Compute the indefinite integral:

\int e^{2 x} d x = \frac{1}{2} e^{2 x} + C

Compute the boundaries:

\int_{- \infty}^{- 1} e^{2 x} d x = \frac{1}{2 e ^{2}} - 0

= \frac{1}{2 e ^{2}} - 0

Simplify

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

\int 6 x^{2} - 3 x^{3} d x

x + y = 7, x = 0, y = 0

- \frac{14}{( t + 15 ) ^{2}}

z^{2} z e^{x yz}

\int tan^{3} (3 π x) d x

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