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

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

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

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

- \frac{6}{e} + 3

Decimal

0.79272 \dots

Solution steps

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

Simplify

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

\frac{5}{x ^{5}} + \frac{3}{x ^{2}}

- 3 x^{2} - 5 x

\frac{d}{d x} (\frac{x ^{2}}{x + 10})

n = 1 \sum \infty \frac{2}{n ^{2}}

y = x (x - 1)^{2}

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