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

The integral from 0 to ln(1) of e^{-t^2} is 0

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integral from 0 to ln(1) of e^{-t^2}

\int_{0}^{l n (1)} e^{- t^{2}} d t

0

Solution steps

\int_{0}^{l n (1)} e^{- t^{2}} d t

Use the nonelementary integral:

\int e^{- t^{2}} d t = \frac{π}{2} erf (t)

= [\frac{π}{2} erf (t)]_{0}^{l n (1)}

Apply log rule:

lo g_{a} (1) = 0

= [\frac{π}{2} erf (t)]_{0}^{0}

Compute the boundaries:

0

= 0

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

What is the integral from 0 to ln(1) of e^{-t^2} ?
The integral from 0 to ln(1) of e^{-t^2} is 0