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

y^{^{''}} + 49 y = 0, y (0) = 4, y^{^{''}} (0) = - 6

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

f (x) = 2 x + 2

\frac{\partial}{\partial x} (arctan (x^{2} y))

\int 5 x x^{2} + 1 d x

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