Answer
a.
the definition of the error function is that
$
\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t
$
$\int_a^be^{-\frac12t^2}dt=\int_0^be^{-\frac12t^2}dt-\int_0^ae^{-\frac12t^2}dt$
So
$\frac{2}{\sqrt\pi}\int_a^be^{-\frac12t^2}dt=\frac{2}{\sqrt\pi}\int_0^be^{-\frac12t^2}dt-\frac{2}{\sqrt\pi}\int_0^ae^{-\frac12t^2}dt$
$\frac{2}{\sqrt\pi}\int_a^be^{-\frac12t^2}dt=\operatorname{erf}(b)-\operatorname{erf}(a)$
So $\int_a^be^{-\frac12t^2}dt=\frac{\sqrt\pi}2[\operatorname{erf}(b)-\operatorname{erf}(a)]$
b.
$\begin{aligned}
y&=e^{x^2}erf(x)\\
&=\frac{2}{\sqrt\pi}e^{x^2}\int_0^xe^{-t^2}dt
\end{aligned}$
$y'=\frac{2}{\sqrt\pi}\left((e^{x^2})'\int_0^xe^{-t^2}dt+e^{x^2}(\int_0^xe^{-t^2}dt)'\right)$
$y'=\frac{2}{\sqrt\pi}(2xe^{x^2}\int_0^xe^{-t^2}dt+e^{x^2}\cdot e^{-x^2}) $
So $y'=2xe^{x^2}(\frac{2}{\sqrt\pi}\int_0^xe^{-t^2}dt+1)=2xy+\frac{2}{\sqrt\pi}$
Work Step by Step
a.
the definition of the error function is that
$
\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t
$
$\int_a^be^{-\frac12t^2}dt=\int_0^be^{-\frac12t^2}dt-\int_0^ae^{-\frac12t^2}dt$
So
$\frac{2}{\sqrt\pi}\int_a^be^{-\frac12t^2}dt=\frac{2}{\sqrt\pi}\int_0^be^{-\frac12t^2}dt-\frac{2}{\sqrt\pi}\int_0^ae^{-\frac12t^2}dt$
$\frac{2}{\sqrt\pi}\int_a^be^{-\frac12t^2}dt=\operatorname{erf}(b)-\operatorname{erf}(a)$
So $\int_a^be^{-\frac12t^2}dt=\frac{\sqrt\pi}2[\operatorname{erf}(b)-\operatorname{erf}(a)]$
b.
$\begin{aligned}
y&=e^{x^2}erf(x)\\
&=\frac{2}{\sqrt\pi}e^{x^2}\int_0^xe^{-t^2}dt
\end{aligned}$
$y'=\frac{2}{\sqrt\pi}\left((e^{x^2})'\int_0^xe^{-t^2}dt+e^{x^2}(\int_0^xe^{-t^2}dt)'\right)$
$y'=\frac{2}{\sqrt\pi}(2xe^{x^2}\int_0^xe^{-t^2}dt+e^{x^2}\cdot e^{-x^2}) $
So $y'=2xe^{x^2}(\frac{2}{\sqrt\pi}\int_0^xe^{-t^2}dt+1)=2xy+\frac{2}{\sqrt\pi}$
