Find the area of manipulated Gaussian function

Question

A unit Gaussian function is defined: $G(x)=\frac{1}{\sqrt \pi} exp(-x^2)$.

The area is unity: 1:

A Gaussian of height a times of the unit Gaussian and width b times that of the unit Gaussian and centred at $x=c$ can be written as:

It as an area of: $ab$

I was wondering how how we integrate the manipulated Gaussian function to find this area?

My working so far:

Answer 1

You have to prove that

$$\int_{-\infty}^{+\infty}\frac{a}{\sqrt{\pi}}e^{-\frac{(x-c)^2}{b^2}}dx=ab$$

let's set

$$y=\frac{x-c}{b}$$

Your integral becomes (reminding what the Gaussian integral is)

$$ab\frac{1}{\sqrt{\pi}}\underbrace{\int_{-\infty}^{+\infty}e^{-y^2}dy}_{=\sqrt{\pi}}=ab$$

Answer 2

A nice way to think of it is that $c$ has no effect on the area, since we are in the domain $x\in(-\infty,\infty)$ so shifting the function, and hence the domain "left or right" will have no effect. because of this we can neglect it so we simply have the integral: $$\frac{a}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-(x/b)^2}dx$$ now using the substitution $u=x/b$ the answer should be pretty clear :)