# Upper Sums

• Mar 1st 2010, 02:38 PM
rawkstar
Upper Sums
use a graphing utility to graph f(x)=x^4-6x^3+11x^2-6x. then use the upper sums to approximate the area of the region in the first quadrant is bounded by f and the x axis using 4 subintervals. round your answer to three decimals

I have no idea how to do this and its due wednesday, we've never even talked about it in class
• Mar 1st 2010, 02:43 PM
icemanfan
First you need to determine the roots of the function, so you know what bounds to use.
• Mar 1st 2010, 02:51 PM
rawkstar
roots = 0,1,2,3
• Mar 1st 2010, 03:13 PM
icemanfan
Quote:

Originally Posted by rawkstar
roots = 0,1,2,3

Well done. Now on which of the intervals $(0, 1), (1, 2), (2, 3)$ is the function positive (so that the area bounded by the function and the x-axis is in the first quadrant)?
• Mar 1st 2010, 03:28 PM
rawkstar
(1,2)
• Mar 1st 2010, 03:36 PM
icemanfan
Quote:

Originally Posted by rawkstar
(1,2)

Good. Now separate that interval into four subintervals $S_1, S_2, S_3, S_4$ of the same length and determine the maximum value of the function $M_1, M_2, M_3, M_4$ on each subinterval. Then the approximation of the area desired is $0.25(M_1 + M_2 + M_3 + M_4)$. Hint: Use the first derivative of the function to determine whether the function is increasing or decreasing on each subinterval.
• Mar 1st 2010, 03:52 PM
rawkstar
ok so i divided it into fourths using the right endpoints and took
0.25[f(5/4)+f(6/4)+f(7/4)+f(2)] and got approximately 0.346
• Mar 1st 2010, 04:20 PM
icemanfan
Quote:

Originally Posted by rawkstar
ok so i divided it into fourths using the right endpoints and took
0.25[f(5/4)+f(6/4)+f(7/4)+f(2)] and got approximately 0.346

This is not the same as the upper sums. For the upper sums, you want the maximum value of the function on each subinterval. It turns out that f(5/4) and f(3/2) are the maximum values of the function on the intervals [1, 5/4] and [5/4, 3/2] respectively, but the other two numbers you have are not.
• Mar 1st 2010, 04:26 PM
rawkstar
how are you supposed to maximize this then
• Mar 1st 2010, 04:45 PM
icemanfan
The maximum value of the function on [3/2, 7/4] is f(3/2). The maximum value of the function on [7/4, 2] is f(7/4). So the sum should be $0.25(f(5/4) + f(3/2) + f(3/2) + f(7/4))$.