Problem

Evaluate the upper and lower sums for $ f(x) = 2 …

10:49

Problem 6 Medium Difficulty

(a) Graph the function $$ f(x) = x - 2 \ln x \hspace{10mm} 1 \le x \le 5 $$
(b) Estimate the area under the graph of $ f $ using four approximating rectangles and taking the sample points to be (i) right endpoints and (ii) midpoints. In each case sketch the curve and the rectangles.
(c) Improve your estimates in part (b) by using eight rectangles.

Answer

a. $1 \leq x \leq 5$
b. (i) $1[ f(2) + f(3)+f(4)+f(5)] $ ; (ii) $1[ f(1.5)+ f(2.5)+ f(3.5)+ f(4.5)]$
c. $0.5[f(1.25)+f(1.75)+\cdots+f(4.75)]$

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Video Transcript

So this problem we have a function F of X is x minus two. Natural log of x. First part was too graphic. So whatever tools you use if your graphing on paper, if you've got your favorite graphing tool, we see the graph here. And what we're asked to do is to use rectangles to approximate the area under the curve. So the first thing in the first part is to use four rectangles. Okay, so if I'm going to use four rectangles, I'm trying to approximate the area that goes from one over 25 So if I use four rectangles, each rectangle is one in length and they say use the right point of each interval to determine the height of that rectangle. So what that means is you're gonna have F F two, f three, f four, and f F five to determine the height of the rectangles. So the first part is going to be the width of the rectangle, so the width Is going to be one and then the height of each rectangle, so the height is gonna be determined by F. Of two. Okay. Plus f. of three Plus f of four Plus F of five. And that will give me my approximation to the area using four rectangles. So let's take a look at this on our calculator. Um So if I store the function, I've stored the function F. Of X. So F. Of X. Sorry about that. Um So I've stored F of X as X minus two natural log of X. So it's one times and then that's going to be F of two Plus f of three F. Of four. But f. of five. And that gives me 4.4-5. So this first estimate with four rectangles right In Points four 4 to 5. So this is about 4.4. And what did I say? 4.4? Uh 25 Yeah. Now they ask is to keep those same for rectangles, but instead of using the right point, use the midpoint, so the mid points will be happening at 1.52 point 53.54 point five. So it's going to be one times, And this is gonna be f of 1.5 Plus f of 2.5 Plus f. of 3.5 Plus f. of 4.5. So that's using the midpoint of each rectangle. So let's just go back and look at that on our calculator. And what I'll come up with there is I can just recall this command And it's just f of 1.5, Half of 2.5 of 3.5 And F of 4.5. So I get the approximation of 3.84, so 3.84 is my approximation yeah. Using mid points. Okay if I want to find out how I'm doing, I looked at this real quick if I looked at the actual value of this Um definite integral is about 3.91 or 3.905. So the exact value is so the the exact value is close To what was that? Um 3.905, So 3.905. So what I see so far is the mid points with four rectangles was better than the right side of each rectangle. What they do in the next part is to say let's do the same exercise but with eight rectangles. So if I have eight rectangles then you've got the 5 -1 is four. When you divide that by eight each rectangle is going to be one half. Okay so let's do that. You're gonna have the width is one half. Yeah? Okay. And then you're going to have eight rectangle. So if I use the right side, okay. So if I split this up using the right side of each one, that's gonna be 1.5 to 2.5 3. So let's try that. That's going to be F of 1.5 Plus f of two Plus f of 2.5 Plus F of three. Okay. Plus f of 3.5 Plus f. A four mm hmm. Plus F 4.5 plus f. Oh, wow of five. Count them up. 12345678 values here. So let's see what this gives us in our calculation. So you're going to have 0.5 times f of 1.5 plus F. Of two Plus f. of 2.5 Plus f. of three plus f. Of Yeah, 3.5 plus f of four Plus f of 4.5 Plus F of five. 4.13 so 4.13 is the approximation that I get there. So this is about 4.13. So what I can see is originally when I did that same calculation with With four rectangles, I got 4.4 with eight rectangles, I got 4.13, so I'm converging closer to the real number, so I should get even a better result if I use the mid points, that was already a better exercise to begin with. So if I look at the mid points, So the mid points are gonna be, so if you look at the rectangle of one half is going to be at the .25, so 1.25 And then 1.75. So that's where I'm going to find. Mid points. So for mid points this is going to be one half. Um So one half and then you're gonna have f at 1.25. Yeah. Plus f of 1.75 Plus F of 2.25. Mhm. Plus f of 2.75 Plus f of 3.25 Plus f of 3.75. Mhm. Plus f of 4.25 Plus f of four point 7512345678 of those. Let's figure out what this one is going to be. So go back to my calculator and let's just change all of these up. So it starts here. Um Yeah, That's 1.25 And then 1.75 And then 2.25, 2.75. Mhm. 3.25, yeah 2.75 4.25. And lastly 4.75 3.889. So my approximation here is 3.889. Yeah. Yeah. Mhm. And you can see that that's even closer that's the best one yet closest to the actual value Of 3.95.

Robert D.

Florida State University

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Problem

Evaluate the upper and lower sums for $ f(x) = 2 …

Problem 6 Medium Difficulty

Answer

a. $1 \leq x \leq 5$
b. (i) $1[ f(2) + f(3)+f(4)+f(5)] $ ; (ii) $1[ f(1.5)+ f(2.5)+ f(3.5)+ f(4.5)]$
c. $0.5[f(1.25)+f(1.75)+\cdots+f(4.75)]$

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a. $1 \leq x \leq 5$b. (i) $1[ f(2) + f(3)+f(4)+f(5)] $ ; (ii) $1[ f(1.5)+ f(2.5)+ f(3.5)+ f(4.5)]$c. $0.5[f(1.25)+f(1.75)+\cdots+f(4.75)]$

Calculus: Early Transcendentals

Grace H.

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Michael J.

Joseph L.

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Grace H.

Kayleah T.

Michael J.

Joseph L.

a. $1 \leq x \leq 5$
b. (i) $1[ f(2) + f(3)+f(4)+f(5)] $ ; (ii) $1[ f(1.5)+ f(2.5)+ f(3.5)+ f(4.5)]$
c. $0.5[f(1.25)+f(1.75)+\cdots+f(4.75)]$