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

The integral from 0 to 1 of e^{-t^2} is 0.74682…

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

\int_{0}^{1} e^{- t^{2}} d t

0.74682 \dots

Solution steps

\int_{0}^{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}^{1}

Compute the boundaries:

0.74682 \dots

= 0.74682 \dots

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 8 y = 0

sec^{2} (2 x + 5) \cdot 2

n \to 2 lim (n^{2})

f (x) = (5 x - 5) (2 x^{3} - x^{2} + 1)

\frac{\partial}{\partial u} (u^{5} + v^{9})

What is the integral from 0 to 1 of e^{-t^2} ?
The integral from 0 to 1 of e^{-t^2} is 0.74682…