A challenging integral

We change the limits to $0$ and $1$ , using $\int_{a}^b f(x).dx=(b-a) \int_{0}^1 f((b-a)x+a).dx$ Now let $I_1=\int_{-4}^{-5} e^{(x+5)^2}.dx$

$\implies I_1=(-5+4) \int_{0}^1 e^{((-5+4)x-4+5)^2}dx.$

$\quad \implies I_1=-\int_{0}^1 e^{(x-1)^2}.dx$ ............................ I

Now let $I_2 =\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}} e^{9.(x-\frac{2}{3})^2} .dx$

$\implies I_2= (\frac{2}{3}-\frac{1}{3}) \displaystyle \int_{0}^1 e^{9. \left( (\frac{2}{3}-\frac{1}{3})x+\frac{1}{3}-\frac{2}{3} \right)^2} .dx$

$\quad \implies I_2= \frac{1}{3} \int_{0}^1 e^{(x-1)^2}.dx$ ..................................... II

Dividig I by II :: $\frac{I_1}{I_2}=\frac{-\int_{0}^1 e^{(x-1)^2}.dx}{\frac{1}{3} \int_{0}^1 e^{(x-1)^2}.dx}=\boxed{-3}$

let us take x+5=y, so LL=1,UL=0 ; and dx=dy and let 9(x-2/3)^2=(3x-2)^2 therefore we can assume 3x-2=t, so LL=-1,UL=0 ;and dx=dt/3 e^x^2 is similar on both sides of y axis, hence numerator and denominator are similar integrals, so it can be manipulated as -I/I/3=-3

We see that both are of form ${e}^{x^2}$ .

In First one Substitute $x + 5 = t$ .

In second one substitute $\frac{2}{3} - x = \frac{t}{3}$

Rest all you can do :)