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

The integral from 0 to x^2 of (1-t)e^{t^2} is (sqrt(pi))/2 \erfi(x^2)-1/2 (e^{x^4}-1)

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

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

\frac{π}{2} erfi (x^{2}) - \frac{1}{2} (e^{x^{4}} - 1)

Solution steps

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

Expand

(1 - t) e^{t^{2}} : e^{t^{2}} - e^{t^{2}} t

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

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

\int_{0}^{x^{2}} e^{t^{2}} t d t = \frac{1}{2} (e^{x^{4}} - 1)

= \frac{π}{2} erfi (x^{2}) - \frac{1}{2} (e^{x^{4}} - 1)

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What is the integral from 0 to x^2 of (1-t)e^{t^2} ?
The integral from 0 to x^2 of (1-t)e^{t^2} is (sqrt(pi))/2 \erfi(x^2)-1/2 (e^{x^4}-1)