What is the integral from 0 to 2 of (x-1)e^x ?

The integral from 0 to 2 of (x-1)e^x is 2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of (x-1)e^x

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

2

Solution steps

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

Apply Integration By Parts

: [e^{x} (x - 1) - \int e^{x} d x]_{0}^{2}

= [e^{x} (x - 1) - \int e^{x} d x]_{0}^{2}

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

= [e^{x} (x - 1) - e^{x}]_{0}^{2}

Expand

e^{x} (x - 1) - e^{x} : e^{x} x - 2 e^{x}

= [e^{x} x - 2 e^{x}]_{0}^{2}

Compute the boundaries:

2

= 2

\int_{- 1}^{1} x^{6} sin (x) cos (x) d x

\int_{- 2}^{3} (x^{2} + 3 x - 1) d x

\int_{e^{9}}^{e^{49}} \frac{1}{x ln ( x )} d x

\int_{2}^{3} (x - 1)^{2} d x

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

What is the integral from 0 to 2 of (x-1)e^x ?
The integral from 0 to 2 of (x-1)e^x is 2