What is the integral from 0 to 1 of (x^2+8)(e^{-x}) ?

The integral from 0 to 1 of (x^2+8)(e^{-x}) is -(13)/e+10

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integral from 0 to 1 of (x^2+8)(e^{-x})

\int_{0}^{1} (x^{2} + 8) (e^{- x}) d x

- \frac{13}{e} + 10

Decimal Notation

5.21756 \dots

Solution steps

\int_{0}^{1} (x^{2} + 8) e^{- x} d x

Apply Integration By Parts

: [- e^{- x} (x^{2} + 8) - \int - 2 e^{- x} x d x]_{0}^{1}

= [- e^{- x} (x^{2} + 8) - \int - 2 e^{- x} x d x]_{0}^{1}

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

= [- e^{- x} (x^{2} + 8) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1}

Simplify

[- e^{- x} (x^{2} + 8) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 10 e^{- x}]_{0}^{1}

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

Compute the boundaries:

- \frac{13}{e} + 10

= - \frac{13}{e} + 10

x \to 2 - lim (\frac{x - 3}{x - 2})

\int \frac{e ^{x}}{e ^{x} - 1} d x

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

\int_{2}^{5} \frac{ln ( x ^{5} )}{x} d x

(1 + x)^{\frac{1}{3}}

What is the integral from 0 to 1 of (x^2+8)(e^{-x}) ?
The integral from 0 to 1 of (x^2+8)(e^{-x}) is -(13)/e+10