What is the integral from 0 to 1 of 4(x^2+1)e^{-x} ?

The integral from 0 to 1 of 4(x^2+1)e^{-x} is 4(-6/e+3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 4(x^2+1)e^{-x}

\int_{0}^{1} 4 (x^{2} + 1) e^{- x} d x

4 (- \frac{6}{e} + 3)

Decimal

3.17089 \dots

Solution steps

\int_{0}^{1} 4 (x^{2} + 1) e^{- x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{1} (x^{2} + 1) e^{- x} d x

Apply Integration By Parts

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

= 4 [- e^{- x} (x^{2} + 1) - \int - 2 e^{- x} x d x]_{0}^{1}

\int - 2 e^{- x} x d x = - 2 (- e^{- x} x - e^{- x})

= 4 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

4 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : 4 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= 4 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= 4 (- \frac{6}{e} + 3)

t \to 0 lim (\frac{e ^{t} - t - 1}{t})

\int_{0}^{2} π (e^{- x^{2}})^{2} d x

f (x) = x^{- \frac{1}{4}}

f (x) = e^{\frac{x}{x - 1}}

e^{x}, at x = ln (7)

What is the integral from 0 to 1 of 4(x^2+1)e^{-x} ?
The integral from 0 to 1 of 4(x^2+1)e^{-x} is 4(-6/e+3)