💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 31 Easy Difficulty

Find the exact area under the cosine curve $ y = \cos x $ from $ x = 0 $ to $ x = b $, where $ 0 \le b \le \pi/2 $. (Use a computer algebra system both to evaluate the sum and compute the limit.) In particular, what is the area if $ b = \pi/2 $?


$\int_{0}^{b} \cos x d x=\lim _{n \rightarrow \infty} \frac{b}{n} \sum_{i=1}^{n} \cos \left(\frac{i b}{n}\right)$

More Answers


You must be signed in to discuss.

Video Transcript

Yeah. Yeah. So problem 5.1. Number 31. So they want us to find the exact area under the cosine curve. So why is he going to cosign X? We're going to go from X equals zero to K. And we're guaranteed the K is between zero And pi over two. So in this case we're going to use rectangles. And so the width of a rectangle is going to be K zero over in because I have in rectangle so Kay over in is going to be the width of every one of these rectangles. Now the other thing, I know I know that this value of K is between zero empire were too. So I know that why equal co sign X. That is greater than equal to zero on the interval 0 to Pi over two. So that means that when I graft this guy And so you should know what the graph of this looks like from zero Pi over to this guy starts out at one ends at zero. So you've got something going on like this, we are going to estimate within rectangles. Okay. In order to and then set the limit to get the exact value. So I see that every rectangle is above the X. Axis. So I'm going to start using right in points. So each in each um each with of each rectangle is K. Or in. And then you're going to have the value of the co sign at that point. So the first one is going to be co sign of ah pi over in and then co sign. Excuse me. Hi over in. And then from there you'll have co sign of. And sorry about that was the case. So it's K. Mhm. Let me back up. So sorry about that. So when you do one, so it's gonna be the co sign of K over in. So co sign of K over in. The next one would be the co sign of twice that two K. Ever in three K. Over in. Until finally you get to um ah um get U. K. Okay so in this case um so let's take a look at it. What would be our some? So the area is going to be the limit as the number of rectangles approaches, infinity of the sum of all the areas of each rectangle. So I'm going to have in rectangles the width of each rectangle is K over in. Okay. And then the height of each rectangle is being, it's gonna be driven by the cosine function. And so if I'm taking the right side input so that's going to be the co sign of K. Or in two K over in. So it's going to be um Kay I over in. So that's the limit that needs to be evaluated in order to get to the value here. So we're going to use um uh some calculating help to make this happen. So if I look at this calculation so the sum yeah I equal end to end of K over in cosign K I over in this one is quite complicated. It's going to take the view so Kay and then sign K. So that's going to be equal to Okay Yeah. Sign Ok. Um coach and urgent care were to in co tangent two K over in Plus co sign K -1 plus co sign. Okay yeah -1 all of that. Mhm Yeah. Over to in. So that is what you see here. That is the sum. So the sum that you see right here is represented by this formula. Okay. So that gives us a some but now we're asked to, since that was a little bit complicated, let's evaluate all of this um using our computer algebra system. So the area is going to be equal to, let's go back now and let's just take the same thing that you see here, take that some. But now let's go find the limit. Mhm. As in approaches infinity. Yeah. So what you see here, it matches what we had the limit as n approaches infinity of the sum of all the rectangles, the width times the height. So that evaluates to the sign of K. So the area is the sign of K. Now they ask us another question. They say what if okay Is equal to Pi over two. If that is the case then you know the area is the sign of the power to which is one. Okay. So what that means is if I look at the coastline curve that goes from zero to pi or two, I now know that the exact area under that curve is one