Some computer algebra systems have commands that will draw...

Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas, at least if $ x_{i}^{*} $ is a left or right endpoint. (For instance, in Maple use leftbox, rightbox, leftsum, and rightsum.) (a) If $ f(x) = 1/(x^2 + 1) $, $ 0 \le x \le 1 $, find the left and right sums for $ n $ = 10, 30, and 50. (b) Illustrate by graphing the rectangles in part (a). (c) Show that the exact area under $ f $ lies between 0.780 and 0.791.

Some computer algebra systems have commands that will draw approximating rectangles and evaluate the sums of their areas, at least if $ x_{i}^{*} $ is a left or right endpoint. (For instance, in Maple use leftbox, rightbox, leftsum, and rightsum.)

(a) If $ f(x) = 1/(x^2 + 1) $, $ 0 \le x \le 1 $, find the left and right sums for $ n $ = 10, 30, and 50.

(b) Illustrate by graphing the rectangles in part (a).

(c) Show that the exact area under $ f $ lies between 0.780 and 0.791.

Transcript

00:02 So problem 5 .11, so we're dealing with area approximation using rectangles and remand sums.

00:11 What we're asked here to do is to find the left and the right sum for this function, f of x is 1 over x squared plus 1 for different numbers of rectangles.

00:21 So if we look at this, a couple of things we know about this function.

00:25 If i look at the first derivative of this function, that it's going to be minus 2x over.

00:33 Over x squared plus 1 squared.

00:36 And on the interval from 0 to 1, this is always negative.

00:43 So the first derivative is always going to be less than or equal to 0.

00:49 That means this function is always decreasing.

00:53 So if a function is always decreasing, and i draw a rectangle, if i draw a left rectangle, that's the same as an upper sum.

01:03 And if i have a rectangle and i use the right, end point, that's going to be a lower sum.

01:09 Okay? so what we're asked to find is using left and right.

01:13 So the left is going to be an upper sum and the right is going to be a lower sum in this particular case because of the decreasing function.

01:22 Now, if we take a look at this graphically, we'll see.

01:25 Here's one over x squared plus one.

01:27 If i use 10 rectangles and i use the left sum, so you see if i use the left side, i'm always getting an upper sum.

01:35 So let's just go ahead.

01:36 We're going to do, let's do all of the left, which are the upper, and then we'll move on to all of the right to do that.

01:42 So if i'm doing all of the left sums, or let's do, i think i'll do, maybe i'll do the case n equal 10, and then we'll move on from there.

01:52 So in the case of n equal 10, let's go ahead and let's do that real quickly.

01:57 So for this, i want to do the function is 1 over x square plus 1, and i want to use 10 rectangles in doing the left sum.

02:07 My function is 1 over x squared plus 1, and that is going to be stored.

02:16 We'll just call this f of x.

02:18 So that is my value of f of x.

02:21 And then now what i want to do is just create the left sides of each of those rectangles.

02:26 So to create those numbers, for the left side, you start with zero.

02:31 If i'm using 10, so the width of the interval is 1.

02:35 So if i'm using 10, i'm going to have the interval.

02:39 The width of being one tenth.

02:40 So i'm going to go from 0 .9, and my increment is going to be 0 .1.

02:46 And i'm going to store this as all of my x values.

02:49 So these are the left side of those rectangles.

02:52 These are my x values.

02:56 And so i have 10x values, which are the left sides of those rectangles.

03:00 Now what i need is just to evaluate the sum.

03:03 So the width of each rectangle is 0 .1.

03:06 And then the height of each rectangle is going to be the function evaluated.

03:11 So it's going to be the sum of all of these x values.

03:18 So 809981.

03:20 So that is the value 809981.

03:24 So if we look at that, 0 .8 .0.

03:37 What was that number again? it was 809981.

03:45 So the width of each rectangle here was one tenth.

03:49 Now let's just do this same thing while i've got the number 10 there.

03:53 If i were to use right endpoints, so in the right endpoints, i don't start at 0 .1.

04:00 So what i would do there is i would go back and say, okay, this sequence that i just created, instead of starting at 0, i'm going to start at 0 .1, and then i'm going to end at the value of 1.

04:14 And those are going to be my new x values.

04:16 And now when i look at the sum, i get 759981.

04:24 So 759981...

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