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

f (x) = cos (\frac{π}{x})

\frac{d}{d x} ((x^{2} - 1) (- 3 x^{3} + 2 x))

\int_{0}^{3} (x + 2) d x

\frac{\partial}{\partial x} (2 x - 2 y)

\frac{d}{d x} (\frac{1 - x ^{2}}{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