If $I_1 =\int_0^xe^{zx}e^{-z^2}dz$ and $I_2 =\int_0^xe^{-z^2/4}dz$, then
Answer & explanation
Correct answer: option 4
$I_1 =\int_0^xe^{zx}e^{-z^2}dz=\int_0^xe^{-(z^2-zx)}dz$
$=\int_0^xe^{-\left\{(z-\frac{x}{2})^2-\frac{x^2}{4}\right\}}dz=e^{x^2/4}\int_0^xe^{(z-\frac{x}{2})^2}dz$
$=e^{x^2/4}\int_{-x/2}^{x/2}e^{-z^2}dz=\frac{1}{2}e^{x^2/4}\int_{-x}^xe^{-z^2/4}dz$
$=e^{x^2/4}\int_0^xe^{-z^2/4}dz=e^{x^2/4}I_2$
Hence, $I_1=e^{x^2/4}I_2$