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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.

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Here is a demo for doing convolution with respect a set of choices of filters. There is an animate button as well.

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I can't find the "animate" button –  Amr Feb 14 '13 at 10:31
    
Next to "1D filters" there are some choices. If you click on any of them the graph of the corresponding filter shows. Next to that graph is the button for "animate". –  Maesumi Feb 14 '13 at 13:25

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