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

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

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

\int_{- 1}^{1} \frac{1}{1 + e ^{- x}} d x

1

Solution steps

\int_{- 1}^{1} \frac{1}{1 + e ^{- x}} d x

\frac{1}{1 + e ^{- x}} = - \frac{e ^{- x}}{1 + e ^{- x}} + 1

= \int_{- 1}^{1} - \frac{e ^{- x}}{1 + e ^{- x}} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{1} \frac{e ^{- x}}{1 + e ^{- x}} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} \frac{e ^{- x}}{1 + e ^{- x}} d x = 1

\int_{- 1}^{1} 1 d x = 2

= - 1 + 2

Simplify

= 1

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