What is the integral from 0 to 1 of-10(x^2+1)e^{-x} ?

The integral from 0 to 1 of-10(x^2+1)e^{-x} is -10(-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-10(x^2+1)e^{-x}

\int_{0}^{1} - 10 (x^{2} + 1) e^{- x} d x

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

Decimal Notation

- 7.92723 \dots

Solution steps

\int_{0}^{1} - 10 (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

= - 10 \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}

= - 10 [- 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})

= - 10 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

- 10 [- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : - 10 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

= - 10 [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= - 10 (- \frac{6}{e} + 3)

\frac{d y}{d x} = \frac{( x ^{2} - y )}{x}

\int x + 2 x d x d x

x \to 0 + lim (x^{2} - x - ln (x))

\int \frac{x ^{3} - 2 x - 2}{x ^{4} + 2 x ^{3} + 3 x ^{2}} d x

\frac{d}{d x} (x^{\frac{4}{5}} (x - 6)^{2})

What is the integral from 0 to 1 of-10(x^2+1)e^{-x} ?
The integral from 0 to 1 of-10(x^2+1)e^{-x} is -10(-6/e+3)