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

The integral from 0 to 8 of (x^2+1)e^{-x} is -(83)/(e^8)+3

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

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

- \frac{83}{e ^{8}} + 3

Decimal Notation

2.97215 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

[- e^{- x} (x^{2} + 1) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{8} : [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{8}

= [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{8}

Compute the boundaries:

- \frac{83}{e ^{8}} + 3

= - \frac{83}{e ^{8}} + 3

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What is the integral from 0 to 8 of (x^2+1)e^{-x} ?
The integral from 0 to 8 of (x^2+1)e^{-x} is -(83)/(e^8)+3