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

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

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

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

1

Solution steps

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

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

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

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 ^{- t}}{1 + e ^{- t}} d t + \int_{- 1}^{1} 1 d t

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

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

= - 1 + 2

Simplify

= 1

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