(a) If $ f(x) = \ln x $, $ 1 \le x \le 4 $, use the commands discussed in Exercise 11 to 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 2.50 and 2.59.
(A). $L 10 \approx 2.331630 R 10 \approx 2.747518 L 30 \approx 2.475238 R 30 \approx 2.613867 L 50 \approx$ 2.503364$R 50 \approx 2.586541$
(B). Click to see
(C). $4 \ln 4-3 \approx 2.54517744$
Mhm. So a problem. 5, 12. We're looking at using rectangles for area approximation. So let's look at the function F of X equals the natural log of X. Yes, On the domain, 1- four. So what do we know about this function? You know that f prime of X is equal to one over x. and on that interval from 124 you know that this is always going to be positive. So this function is always increasing over its domain. So that's going to help us because if a function is increasing over its domain, if you use a right side rectangle, so right side rectangle is going to give you an upper some. A left side rectangle is going to give you a lower some. So that's going to help us as we look at this. So what we're asked to do in this particular case is to say, Let's take a sampling of a number of rectangles, 10, 30 And 50 that will dictate the width of each rectangle. So if I go over the interval 1-4 so 4 -1 is going to be equal to three. So the width here is going to be 3/10 here. And then the width you see in the next one is 3/30. Which is also yeah 1/10. Yes. And then three 50th. Those are going to be the wits of my rectangles. And so we'll take the case where we do a left some and a right some. Now if we take a look at this um here's an example of in equal 10 and this is a left some. So you can see very clearly that if I use the left side of each of those rectangles, I'm getting a lower some. So this is going to give me my lower some. And then when you look at that same drawing, if I were to change and use the right side of each rectangle. So if I use the right side now you can see when I use the right side of each rectangle, I'm getting an upper some. So the right side is going to give me um my upper some, so the right side will be my upper some. And so let's just go ahead and start doing this with our calculator. So function was natural log of X. So as we get this in, let's go to our calculator, let's input the function first. So the function is natural log of X. And I'm going to store that. We're going to call that f of X. Now let's go ahead and generate the sequence of all the in points. Uh We'll do the left and then we'll do the right. So so I'm going to sequence all of these numbers if I start on the left So remember this is going from 1-4. So I start on the left, I'm starting at the number one And so if I start at the number one then where do I go from there? Let me just look at the syntax real quick. Um Yeah so start at the number one and then I'm going up to on the left side. If each with is 1/10 It's going to be 4 -1/10. So that's going to be 3.9. Go up by .1 and we're going to store those as my ex values. So this should give me all of the X. Values. Um And if I look at that list that starts at one on the left and it should end At 3.9 which it does. Okay so that is the most set of all the X values when I have the um um The width of 1/10. Okay um and let me back up. Sorry that was wrong, it was the width should be three tens so let me go back and correct that. So the width is going to be 3/10, So I'm starting at one and then if the width is 3/10 that's gonna be four minus 3/10. So that's 3.7, so 3.7 and instead of this being 1/10 this is going to be pretence. That should give me the 10 values. Sorry about that. So this is my in equal 10 scenario now to figure out the sum of all of the areas where the width of each rectangle is 3/10 times the height of each rectangle is going to be the function natural log of X evaluated at the left end point. So that's going to be the sum of all of the X values. So 2.331. Uh we could just call it 2.3316. So 2.3316. Yeah that's my left some and that is my lower some. So now let's do the same thing with that in equal temples. Do write some. If the right some upper then I expect that to be a number that is higher. So let's go to the right some the only difference if I look at the sequence okay the right some is not going to start at one, it's going to start at one plus 3/10 so 1.3 And then it's going to end at four on the very right side boundary and it's still stepping by three tens. And so now um the height is gonna be the same value. So 2.74. So 274752 2.74752. And as expected the upper summit is higher than the lower some. Now let's move on to the case where N. is equal to 30th, it is equal to 30. So if I go back to my calculator and now I'm going to recreate that sequence and let's just do it from the left. Um So I'm starting at one still. Um but now if N is equal to 30 so keep in mind when N was equal to 30 That if I look at my width the width is 1/10 So go back to my calculator. The width here is going to be 1/10. So if the wits is 1/10 my last one is going to end at 3.9. This gives me all the values of the 30 left sides. And then you find that okay if the width is 1/10 and one divided by 10 times the sum of all of those X values. Yeah it's my approximation here Which is 2475-4 so 2475- four. Mhm. Yeah that is my left lower some Now let's repeat this same process when N is equal to 30 but let's do a write up or some. So go back to my calculator and let's just create this on the um from creating this on the right side. I don't start at one I start at 1.1 because the width is 1/10 and I don't end at 3.9. I ended four. Yeah That is my list of um 30 x values Now then you would go and say well the width was 1/10. Actually I've got that same command right here. Um to evaluate that I get 2.61 387 So 261387. And in each case you're finding the upper some to be um higher than the lower some in this particular case. Now last case let's go to end is 50. So to go for an equal 50 back to my calculator. Yeah and let's just go do the left sign. But when N is 50 the width of that is going to be 3/50. Yeah. Yeah the width of each rectangle. So I started one and then you're going to have 3/50 is what? Six over 100? Um So that's going to be um 3.94 or we could just say it's going to be the end boundary. What, four 3/50. And that should give me all of the x rays and if I see that then the X value should end After I get all 50 of those values and that should be fine. Let's just make sure it ends where we want it slightly shy of four. Yes. So that is slightly show for Um yeah, so 3.94. So that gives me the values for the sequence and then I'll figure out okay, the width here was 3/50 so three divided by 50 times the sum of all of the X values effort. All of the X values. So 2.50336. 2.5336. Yes mm And now we just go back and repeat that same process. But now I'm looking at the right side so if I'm looking at right side rectangles, let's go back to our sequence. Instead of starting at one it's one plus 3/50. That's where I'm starting and I'm going to end at four the right side. That gives me all of my ex values. And so now you just do the same command The 50th times the sum of all those x values. 2.58654. So 258 654 Yeah. So what you see here is that now this number right here When energy equal to 50 this is my best approximation. So this is my best lower estimate. And then you also have your best upper estimate. Yes. So you know that the actual value. Yeah. Yeah the actual value has to be in between. So the actual value has to be between these two numbers. Because if you've got a lower estimate then your actual value has to be higher than your lower estimation and has to be lower than your upper estimation. Because each of these, an upper is going to overshoot the right value, the correct value and a lower estimate is going to, worst case is going to undershoot the value. So somewhere the actual value has to be in between these two numbers.