MAT 156 sp'09 Lab 3 More on Riemann Sums and Integrals You should not do this lab until you have completed Lab 2. You can add color to your leftbox command as follows. with(student): with(plots): leftbox(x^2,x=0..4,5,shading=yellow);

Now I am going to combine two leftsum pictures with different colors. lb10:=leftbox(x^2,x=0..4,10,shading=yellow): lb5:=leftbox(x^2,x=0..4,5,shading=white): display(lb5,lb10);

You can easily see the difference between the two left sums now. (Warning: Maple seems tempermental about these commands. For example writing display(lb10,lb5) instead of display(lb5,lb10) matters. So be patient and experiment wtih the commands.) The problem is that the first plot covers the second and the second does not show through. Assignment Part 1: (Do not forget the plots and student packages). Using sin(x) and x=0..Pi/4. A) Graph the left-hands sums with 5 and 10 subintervals simultaneously. Explain why LHS(5) is less than LHS(10). B) Graph and explain why RHS(5)>RHS(10)>RHS(20). Assignment Part 2: Using sin(x) and x=0..Pi/2. Write commands to compute the left-hand sums and the right-hand sums with 10, 20, 40, 80, 160 subintervals. We see that indeed the left-hand sums are increasing and the right-hand sums are decreasing. At the same time the right-hand sums are always larger than all the left-hand sums. As you see we have not gotten great accuracy. To compute more accurately it will be nice to increase the number of subintervals. As it is tiring to write the same commands over and over, it is important to introduce loops in Maple. Loops and convergence of Riemann sums. The following commands produce a table of the left-hand sums with 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120 subintervals. Notice that every time we multiply the number of subintervals by 2, so at the first step (j=0) we have 5*2^0 subintervals, at the second step (j=1) we have 5*2^1=10 subintervals, at the third step (j=2) we have 5*2^2=20 subintervals etc. This gives the formula 5*2^{j} in the commands. for j from 0 to 10 do n:= 5*2^(j): evalf(leftsum(x^2, x=0..4, n)):od; We see that the first two decimal places have stabilized. However, as you have noticed before, the left-hand sums are increasing. To be certain that the second decimal place is indeed 2 we need to give overestimates, which are the right-hand sums. Assignment Part 3: Write commands that produce the right-hand sums with the same number of subintervals, using a loop. All you have to do is copy the command for the left-hand sums and change leftsum into rightsum. Computing Integrals with Maple int(x^2,x=0..4); Assignment Part 4: Study 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 as follows: A) Compute the left-hand sums and right-hand sums with 2, 4, 8, 16, 32, 64, 128, 256, 516, 1024 subintervals. Use a loop. B) How many decimal digits are you certain of? C) Which are overestimates and which are underestimates and why? D) Do the left-hand sums form an increasing or decreasing sequence of numbers, as the number of subintervals increase? Show this in appropriate graphs. E) Compute the exact integral.