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

The integral from 0 to x of (e^{t^2}-1) is (sqrt(pi))/2 \erfi(x)-x

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

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

\frac{π}{2} erfi (x) - x

Solution steps

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

\int_{0}^{x} e^{t^{2}} d t = \frac{π}{2} erfi (x)

\int_{0}^{x} 1 d t = x

= \frac{π}{2} erfi (x) - x

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What is the integral from 0 to x of (e^{t^2}-1) ?
The integral from 0 to x of (e^{t^2}-1) is (sqrt(pi))/2 \erfi(x)-x