What is the integral from 0 to 5 of xe^{x-5} ?

The integral from 0 to 5 of xe^{x-5} is 4+1/(e^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 5 of xe^{x-5}

\int_{0}^{5} x e^{x - 5} d x

4 + \frac{1}{e ^{5}}

Solution steps

\int_{0}^{5} x e^{x - 5} d x

Apply Integration By Parts

: [e^{x - 5} x - \int e^{x - 5} d x]_{0}^{5}

= [e^{x - 5} x - \int e^{x - 5} d x]_{0}^{5}

\int e^{x - 5} d x = e^{x - 5}

= [e^{x - 5} x - e^{x - 5}]_{0}^{5}

Compute the boundaries:

4 + \frac{1}{e ^{5}}

= 4 + \frac{1}{e ^{5}}

\int_{- 1}^{0} x^{5} d x

\int_{- c}^{c} \frac{3}{2} x^{2} d x

\int_{1}^{2} x 6^{(x^{2})} d x

\int_{- 1}^{0} 3 t^{2} e^{- t^{3 + 1}} d t

\int_{0}^{π} 18 cos (6 x) + 6 y z^{2} d x

What is the integral from 0 to 5 of xe^{x-5} ?
The integral from 0 to 5 of xe^{x-5} is 4+1/(e^5)