I don't exactly remember whether I should get the common area under the curves of the functions being convolved or I should multiply them and get the area under the resulting curve.
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The formula for convolution of $f$ and $g$ is $(f \star g)(x) = \int f(s) g(x-s)\ ds$. So you take the area under the product of one function and a reflected and translated version of the other.
Here is a demo for doing convolution with respect a set of choices of filters. There is an animate button as well.