What is the integral from 0 to x of e^{5t^2} ?

The integral from 0 to x of e^{5t^2} is (sqrt(pi))/(2sqrt(5))\erfi(sqrt(5)x)

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integral from 0 to x of e^{5t^2}

\int_{0}^{x} e^{5 t^{2}} d t

\frac{π}{2 5} erfi (5 x)

Solution steps

\int_{0}^{x} e^{5 t^{2}} d t

= \int_{0}^{x} e^{(5 t)^{2}} d t

Apply u-substitution

: \int_{0}^{5 x} e^{u^{2}} \frac{1}{5} d u

= \int_{0}^{5 x} e^{u^{2}} \frac{1}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int_{0}^{5 x} e^{u^{2}} d u

Use the nonelementary integral:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= \frac{1}{5} [\frac{π}{2} erfi (u)]_{0}^{5 x}

Compute the boundaries:

\frac{π}{2} erfi (5 x)

= \frac{1}{5} \cdot \frac{π}{2} erfi (5 x)

Simplify

= \frac{π}{2 5} erfi (5 x)

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What is the integral from 0 to x of e^{5t^2} ?
The integral from 0 to x of e^{5t^2} is (sqrt(pi))/(2sqrt(5))\erfi(sqrt(5)x)