00:01
For this problem, we're going to use the convolution theorem shown down here.
00:06
And we're going to write our integral here as the convolution of two functions.
00:10
Or if we have one, f -tilda, which is equal to t squared, and another g, which is equal to e to the t.
00:20
Then by this definition, f -tilda, the convolution of f -tilda and g, will therefore be the integral from zero to t of f -tilda, which is t squared, so t minus tau squared, times g of ttilda of tau, which is just e to the tau, e tau.
00:43
And this is what we have for our initial integral.
00:48
So therefore, this integral is just the convolution of ftilda and g...