Here is the summary of the integral calculation that we did in class today: To calculate the integral of a function exp(-x^2) from 0 to L, we found the Taylor series for this function (sum for n from 0 to infinity of (-1)^n x^(2n) / n! ), and integrated it term by term from 0 to L, which yields a sum for n from 0 to infinity of (-1)^n L^(2n+1) / (2n+1) / n! To calculate the result with "m" number of terms in the series, create the following m-file (call it "MyIntegral.m", for instance) function I = MyIntegral(m) I = 0; L = 1; fprintf(' Order: Integral value:\n'); for n = 0 : m an = (-1)^n * L^(2*n+1) / (2*n+1) / factorial(n); I = I + an; fprintf(' m = %2d I = %9.6g \n', n, I) end Now when you enter a command "MyIntegral(7)", you will see the result approach the true value of the integral (which is equal to 0.746824133), as the number of terms in the Taylor polynomial approximation is increased from 0 to 7. Please make sure you understand exactly what each line in this program does, and let me know if you have any questions. We will eventually move to more complicated programs, so you should feel comfortable with short programs like this one, involving a "for" cycle. Note that you can make the upper limit of integration L to serve as the 2nd argument of your function "MyIntegral", instead of fixing the value of L in the body of the m-file. This will make this function more useful (more flexible). To do this, simply replace the first command with "function I = MyIntegral(m,L)"