Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Estimating the area under a curve can be done by adding areas of rectangles. The area of a rectangle is A=hw, where h is height and w is width. To find the width, divide the area being integrated by the number of rectangles n (so, if finding the area under a curve from x=0 to x=6, w = 6-0/n = 6/n.
The height of the rectangle will be f(a) at whatever number a the rectangle is starting. This depends on whether you are taking a left-hand or right-hand sum. For a left hand sum, the height will be taken from the leftmost part of the rectangle. So, if starting at x=0, the height of the first rectangle will be f(0). The height of the next rectangle will be f(0+w), and so on. For a right hand sum, the height will first be taken from the right side of the rectangle. Going back to the example, the height of the first rectangle in a right-hand sum will be f(0+w).
The area under the curve can be approximated by adding the areas of the rectangles. The left-hand and right-hand sums may be different.
Let's do a problem where we estimate the area under a curve using rectangular sums. Here the problem says estimate the area under the curve f(x) equals 12 over x, from x equals 1 to x equals 4 with left, and right hand sums for n equals 3. So n equals means three rectangles.
I've taken liberty of drawing a picture here. I'm going to do the left hand sum first. You can tell that this is a left hand sum because these rectangles reach up and touch the curve in your upper left corners. So all I have to do, is figure out what the area of each rectangle is and I add them up.
So our left hand sum which I'll abbreviate as LHS, is going to be this height times this width, this height times this width, and this height times this width. Each of the widths is going to be the same. Actually I need to calculate that first. The width I will call delta x. Delta x is going to be this width, which is 4 minus 1, divided by the number of rectangles which is 3. This is just 1. So the first rectangle is going to be f(1).
You going to use the left end point of this sub interval. This subinterval here at the left endpoint. F(1) times the width (I'll fill in the width in a second) plus, the second one is going to be f(2) this is 2, this is 3. F(2) times the width plus f(3) times the width for the third one. This is 3. The height here is f(3).
Now I can factor out the delta xs. It turns out they don't matter anyway, because delta x in this problem was 1. Then f(1) is 12 over 1. F(2) is 12 over 2. F(3) is 12 over 3 times my delta x which is 1. I just have to add these up; 12, 6, and 4, that's 22. So this left hand sum is 22. It's an overestimate of the area under this curve from 1 to 4. Let's take a look at the right hand sum. I'll try to abbreviate RHS.
I need the same delta x. Delta x is 1 in this problem, so this is 2, this is 3. Notice in a right hand sum, the rectangles reach up, and touch the curve on the upper right corner. So for my first rectangle, I'm going to use a height f(2) times delta x. The second rectangle the height is f(3) 3 times delta x. The last rectangle is f(4) times delta x. I'll factor the delta x's out. F(2) is 12 over 2. F(3) is 12 over 3. F(4) is 12 over 4 times the delta x which is 1. So I have 6 plus 4 plus 3. This is 13.
So this gives me an approximation for the area under the curve between x equals 1, and x equals 4, 13. You can see that it is an underestimate. So, if I were to use these two estimates together, I would say that the area under the curve was less than 22, but greater than 13.
Unit
The Definite Integral