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

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

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

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

- \frac{6}{e} + 3

Decimal

0.79272 \dots

Solution steps

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

Apply Integration By Parts

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

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

Simplify

[- e^{- x} (1 + x^{2}) - (- 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

y = \frac{7}{3 - 4 x ^{4}}

(x + 1) y^{^{'}} + (2 x - 1) y = e^{- 2 x}

\frac{d}{d x} (- 6 sin (6 x))

\frac{\partial}{\partial x} (ln (cos (x^{2} y^{2} + z)))

arcsin (\frac{x}{7}), at x = 4

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