My question here is: to calculate the surface of area of a max circle which we can draw between the two functions as shown in the below image and also the surface of the area out this circle.
For the first question: I have calculated the integral $\int_{-1}^{1}2e^{-2x²}dx\,$ which is approximate to $2.395$ but i don't think that is true.
Hint 1...try the following:
Let $(x_0,y_0)$ be the point on the curve $y=e^{-x^2}$ where the normal to the curve passes through the origin.
Then the area of the circle is $\pi(x_0^2+y_0^2)$
Hint 2... To get the area enclosed between the curves and outside the circle, you can adapt the well known result $$\int_{-\infty}^{\infty}e^{-\frac 12x^2}dx=\sqrt{2\pi}$$ to get the area between the curves and then...