Why is convolution the integral of product of two functions

convolutionintegration

I am studying the convolution operation and I have been told that when we calculate the convolution of two functions $f(x)$ and $g(x)$, we are actually calculate "the area under the function $f(\tau)$ weighted by the weighting function $g(t-\tau)$". However, I could not understand why the formula of convolution is $\int_{-\infty}^{\infty} f(\tau)g(t-\tau) \,dx$. To be specific, I don't know why the product of the two functions are calculating the area under the two function.

For example, the area under these two functions

first pic

Should be the same as the area under these two functions

enter image description here

But the product of these two functions are scaled by 1.5. Why does the integral product of these two functions represent the area?

Best Answer

Because the convolution isn't “the area under two functions”.

The convolution, as you cited, can be understood as an area under a function weighted by another function. The word “weighted” is what describes multiplication.

The area under a blue function consists of 3 areas. Orange function defines the weight (density) of those areas: (1) and (3) has density 0, (2) has density 1.5, so the result is: $$ C = 1×0 + 1×1.5 + 1×0 = 1.5 $$

enter image description here