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Q. Let $I _1=\int\limits_0^{ x } e ^{ tx } \cdot e ^{- t ^2} dt$ and $I _2=\int\limits_0^{ x } e ^{- t ^2 / 4} dt$ where $x >0$ then the value of $\frac{ I _1}{ I _2}$ is

Integrals

Solution:

$I_1=\int\limits_0^x e^{t(x-t)} d t $ $t(x-t)=-\left[t^2-t x\right] =-\left[\left(t-\frac{x}{2}\right)^2-\frac{x^2}{4}\right]=\frac{x^2}{4}-\left(t-\frac{x}{2}\right)^2 $ $I_1=\int\limits_0^x e^{\frac{x^2}{4}\left(t-\frac{x}{2}\right)^2} d t=e^{\frac{x^2}{4}} \int\limits_0^x e^{-\frac{(2 t-x)^2}{4}} d t ; 2 t-x=y \Rightarrow d t=\frac{d y}{2}$ $=\frac{e^{x^2 / 4}}{2} \int\limits_{-x}^x e^{-\frac{y^2}{4}} d y$ $I_1=e^{\frac{x^2}{4}} \int\limits_0^x e^{-\frac{t^2}{4}} d t=e^{\frac{x^2}{4}} \cdot I_2 \Rightarrow \frac{I_1}{I_2}=e^{\frac{x^2}{4}}$