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

The integral from 0 to 1 of (x^2+2)e^{-x} is -7/e+4

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

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

- \frac{7}{e} + 4

Decimal

1.42484 \dots

Solution steps

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

Apply Integration By Parts

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

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

Simplify

[- e^{- x} (x^{2} + 2) - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{1} : [- e^{- x} x^{2} - 2 e^{- x} x - 4 e^{- x}]_{0}^{1}

= [- e^{- x} x^{2} - 2 e^{- x} x - 4 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{7}{e} + 4

= - \frac{7}{e} + 4

\frac{d y}{d x} = 9 y e^{x + 5}

y^{^{'}} = x sin (2 y)

\int_{2}^{5} e^{2 x} d x

\frac{\partial}{\partial x} (4 x^{5} - 2 y^{- 3})

\frac{9 + 2 x - 6 x}{x}

What is the integral from 0 to 1 of (x^2+2)e^{-x} ?
The integral from 0 to 1 of (x^2+2)e^{-x} is -7/e+4