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

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

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

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

- \frac{3}{e} + 2

Decimal

0.89636 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - e^{- x} d x = e^{- x}

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

Expand

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

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

Compute the boundaries:

- \frac{3}{e} + 2

= - \frac{3}{e} + 2

\int_{0}^{1} x 1 - x d x

\int_{0}^{2} r e^{- r^{2}} d r

\int_{0}^{π} - 2 sin (x) d x

\int_{1}^{3} \frac{15 + 2 x ^{3}}{x ^{4}} d x

\int_{0}^{2} 1 - ∣ x - 1 ∣ d x

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