Problem

Estimate the area between the graph of the functi…

02:51

Problem 6 Medium Difficulty

Estimate the area between the graph of the function $f$ and the interval $[a, b] .$ Use an approximation scheme with $n$ rectangles similar to our treatment of $f(x)=x^{2}$ in this section. If your calculating utility will perform automatic summations, estimate the specified area using $n=10,50,$ and 100 rectangles. Otherwise, estimate this area using $n=2,5,$ and 10 rectangles.
$$
f(x)=\cos x ;[a, b]=[-\pi / 2, \pi / 2]
$$

Answer

$A_{2}=1.570796, \quad A_{5}=1.933765, \quad A_{10}=1.98353$

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Calculus 2 / BC

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Chapter 5

INTEGRATION

Section 1

An Overview of the Area Problem

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Video Transcript

So this problem we're looking at here is we have a function f of X is equal to the co sign of X and we're working on the interval from Negative Pi over two, two pi over two. So again, I'm gonna be using two rectangles, five rectangles, 10 rectangles. The more rectangles that I use, the closer my approximation will become. Now, my starting point is that negative pi over to my ending point Is that pi over two and with two rectangles, I'm splitting this into two pieces. Now again, this distance from here. Right, So from start to finish would be one pie. So that would mean that the distance from here to here would be pi over two. So that means that my width of each of the rectangles will be pi over two. And I want to add the the some of the outputs in order to get the total area of the rectangles, which will be the some from 1 to 2, because there's only two rectangles of in this case, the co sign. That's my function. Now we're starting at pot negative pi over two, and because and we're really adding hi and over to. We're adding pi over twos to get all of the different rectangles. When I evaluate this, I'm going to get an approximation of the area of 1.57708 This is an approximation from the area only using two rectangles when I switch it over to get five rectangles. Now I'm taking that distance and I'm dividing it into five pieces. Right? So the total distances pie. So each of the wits would be pi over five. And I need the outputs at each one of these values. So that's gonna be the some There's gonna be five rectangles from 1 to 5. The co sign again. I'm starting at negative pie or two, but in this case I'm adding pie and over five each time one pie, one pie over 52 pi over 53 pirate 54 pirate 5555 until he gets a one full unit, which is the distance from start to finish, which is pie. When I plug this into my calculator, I'm going to get an approximation of 1.9338 again getting closer to the actual value. I can repeat this process for 10 units again, the with would be pi over 10 because the total distances pie And if I divided by 10 pi over 10 would be the with the outputs at At each of the intervals there be 10 intervals from 1 to 10 and it would be the co sign again. I'm starting at negative pi over two. But now I'm adding pi over tens. One pi over one pi over 10 to 5 Overton, etcetera, etcetera. And in this case, when I have given approximation, I'm gonna get 1.98 35 The more rectangle they used, the closer my approximation will become. And the only thing that will change in the formula would be these tens. This 10 would become however many rectangles. However many rectangles, however many rectangles

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