What is the integral from-infinity to-5 of θe^θ ?

The integral from-infinity to-5 of θe^θ is -6/(e^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-infinity to-5 of θe^θ

\int_{- \infty}^{- 5} θ e^{θ} d θ

- \frac{6}{e ^{5}}

Decimal Notation

- 0.04042 \dots

Solution steps

\int_{- \infty}^{- 5} θ e^{θ} d θ

Compute the indefinite integral:

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

Compute the boundaries:

\int_{- \infty}^{- 5} θ e^{θ} d θ = - \frac{6}{e ^{5}} - 0

= - \frac{6}{e ^{5}} - 0

Simplify

= - \frac{6}{e ^{5}}

\int_{0}^{1.5} \frac{1}{8} x d x

\int_{0}^{1} 2 cos (x) d x

\int_{9}^{10} 1.5 x^{- 2.5} d x

\int_{2}^{4} - 6 x d x

\int_{0}^{2 π} 2 cos (2 x + 4) d x

What is the integral from-infinity to-5 of θe^θ ?
The integral from-infinity to-5 of θe^θ is -6/(e^5)