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

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

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

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

2 (- \frac{2}{e} + 1)

Decimal Notation

0.52848 \dots

Solution steps

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

Eliminate Absolutes

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

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

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

= 0 + 2 (- \frac{2}{e} + 1)

Simplify

= 2 (- \frac{2}{e} + 1)

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