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

The integral from 0 to 8 of (x^2+8)e^{-x} is -(90)/(e^8)+10

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

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

- \frac{90}{e ^{8}} + 10

Decimal Notation

9.96980 \dots

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

= [- e^{- x} x^{2} - 2 e^{- x} x - 10 e^{- x}]_{0}^{8}

Compute the boundaries:

- \frac{90}{e ^{8}} + 10

= - \frac{90}{e ^{8}} + 10

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

\frac{d}{d x} (x^{4} + 4 x^{2} (y - 2) + 5 (y - 1)^{2})

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

\frac{d}{d x} ((- 2 x^{3} - 4)^{- 4})

f (x) = \frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}}

What is the integral from 0 to 8 of (x^2+8)e^{-x} ?
The integral from 0 to 8 of (x^2+8)e^{-x} is -(90)/(e^8)+10