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

The integral from 0 to 1 of (x^2+x)e^{-2x} is -2/(e^2)+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Decimal

0.22932 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

Expand

- \frac{1}{2} e^{- 2 x} (x^{2} + x) - (\frac{1}{2} e^{- 2 x} x + \frac{1}{2} e^{- 2 x}) : - \frac{1}{2} e^{- 2 x} x^{2} - e^{- 2 x} x - \frac{1}{2} e^{- 2 x}

= [- \frac{1}{2} e^{- 2 x} x^{2} - e^{- 2 x} x - \frac{1}{2} e^{- 2 x}]_{0}^{1}

Compute the boundaries:

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

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

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