Answer
$\int_{a}^{b} e^{-t^{2}}dt=\frac{1}{2}\sqrt \pi[ erf(b)- erf(a)]$
Work Step by Step
As we are given that $erf(x)=\frac{2}{\sqrt \pi} \int_{0}^{x} e^{-t^{2}} dt$
$\int_{0}^{x} e^{-t^{2}} dt=erf(x)\frac{\sqrt \pi}{2} $
Consider $\int_{a}^{b} e^{-t^{2}}dt=\int_{0}^{b} e^{-t^{2}} dt-\int_{0}^{a} e^{-t^{2}} dt $
Therefore,$\int_{0}^{b} e^{-t^{2}} dt-\int_{0}^{a} e^{-t^{2}} dt =[ \frac{\sqrt \pi}{2}erf(b)-\frac{\sqrt \pi}{2} erf(a)]$
Hence, $\int_{a}^{b} e^{-t^{2}}dt=\frac{1}{2}\sqrt \pi[ erf(b)- erf(a)]$
