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

The integral from 0 to 1 of (x^2+1)e^{-x} is -6/e+3

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

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

- \frac{6}{e} + 3

Decimal Notation

0.79272 \dots

Solution steps

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

Apply Integration By Parts

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

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

Simplify

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

= [- e^{- x} x^{2} - 2 e^{- x} x - 3 e^{- x}]_{0}^{1}

Compute the boundaries:

- \frac{6}{e} + 3

= - \frac{6}{e} + 3

n = 0 \sum \infty \frac{2}{5 ^{n}}

\frac{d}{d x} (csc (2 x))

\frac{d y}{d x} = - \frac{x}{4 y}

\frac{\partial}{\partial y} (8 x y^{2} z^{3})

\int_{0}^{2} sin (x) d x

What is the integral from 0 to 1 of (x^2+1)e^{-x} ?
The integral from 0 to 1 of (x^2+1)e^{-x} is -6/e+3