Write the convolution integral in its standard form. This form is specifically the integral of two functions: f and g. The inside of the integral should be the function f(t-b) multiplied by the function g(b). Here, "t" is the upper limit of the integral (the lower limit is zero) and "b" is the variable we are integrating over. In mathematical notation, the convolution integral's standard form is written h(t) = int[f(t-b)g(b)db] where b goes from 0 to t. For example, if the function f(x) = x and g(x) = exp(x) where "exp" is the exponential function, then the convolution integral is h(t) = int[(t-b)exp(b)db].
Separate the convolution integral into two integrals. Note that the function f is additive (it is adding two variables). Use algebra to reverse the factoring of fg. Take the initial integral over the two separate functions. For example, if the convolution integral is int[(t-b)exp(b)db], we can rearrange fg, which is (t-b)exp(b), to a new additive function: texp(b) -- bexp(b). Taking the integral over this new function allows us to separate the integral into two parts (separating where the addition or subtraction sign is. The result for this example is int(texp(b)db) -- int(bexp(b)db).
Compute the individual integrals. Find the integrals separately, in accordance with normal calculus. For the example, the first integral, int(texp(b)), evaluates to texp(b). The second integral, int(bexp(b)db), evaluates to bexp(b) -- exp(b).
Evaluate the solutions to the integrals from zero to t. Plug t into b and and then subtract the value of the integral when you plug 0 into b. The first integral in our example becomes texp(t) -- t. The second integral becomes texp(t) -- t -- exp(t) + 1.
Subtract the second evaluated solution from the first to find the convolution integral. Simply if needed. The example's final solution is texp(t) -- t -- [texp(t) -- t -- exp(t) + 1]. Simplying yields exp(t) + 1.