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Problem 1.(5 points) Find Rg the Regular Right Sum with 3 rectangles in order t0 approximate J5.5 2r? dr.5.5 -Since @ =4,6=5.5, and n = 3, thus W = (6 = a)nList th...

Question

Problem 1.(5 points) Find Rg the Regular Right Sum with 3 rectangles in order t0 approximate J5.5 2r? dr.5.5 -Since @ =4,6=5.5, and n = 3, thus W = (6 = a)nList the test points I;a + iw that we will use for our 3 rectangles:List the heights of our 3 rectangles: f(c;)List the areas of our 3 rectangles wf(c;)Sum the aroas: R3 5e wf(r;)Note: You can earn partial credit on1 this problom

Problem 1. (5 points) Find Rg the Regular Right Sum with 3 rectangles in order t0 approximate J5.5 2r? dr. 5.5 - Since @ =4,6=5.5, and n = 3, thus W = (6 = a)n List the test points I; a + iw that we will use for our 3 rectangles: List the heights of our 3 rectangles: f(c;) List the areas of our 3 rectangles wf(c;) Sum the aroas: R3 5e wf(r;) Note: You can earn partial credit on1 this problom



Answers

If $ R = [0, 4] \times [-1, 2] $, use a Riemann sum with $ m = 2 $, $ n = 3 $ to estimate the value of $ \iint_R (1 - xy^2)\ dA $. Take the sample points to be (a) the lower right corners and (b) the upper left corners of the rectangles.

So we're gonna try and estimate the area under this curve. Why is equal to three X. Um bounded by the curve and the X. Axis um from X. Is equal to 02 X. Is equal to three. And the first way we're gonna do is we're gonna have an equaling Um three. So we're gonna have three different rectangles of equal size. So our delta X. In this case it's going to equal One since we're going from X is equal to 02, X. is equal to three. So if we wanted to look at this graphically really quick just to get an idea of what we're doing. You have the curve here, I'm gonna put in red, This should be Y. is equal to three X. And then we have X. is equal to three to end one. So what we're going to have is we're gonna have three rectangles here. Um The first one is going to be when X. Is equal to zero. The second one, no first one is gonna be the next equal to zero. The second one is gonna be when X. Is equal to 10 and then the third one is sorry when X. Is equal to one. And then the third one is going to be when X is equal to two. So these are gonna be our three rectangles here since we want our rectangles under the curve. Um So that would be a visual way to look at this. So we're going to have our area is going to be equal to our delta X. Which is one times are Y values at the three X. Values we're looking at and we're looking at X. Is equal to zero. So that's zero plus X. Equal of one that's equal to three and then next equal to two, Y. Is equal to six. So this would be equal to one times nine which is just nine. Yeah and now we're going to find the area when n. Is equal to 10. So we're looking at 10 rectangles now so our delta X. Is going to be equal 2.3. We just divide to find the delta X. We just divide the total width which in this case we're going from X. Is equal to zero 2 3 star a total width would be three. We divided by N. three divided by 10 is 3/10 or .3. So now our area A is going to be equal to adult X. Which is 0.3. Then multiplied by our 10 wreck tangled. Um By the 10 Y values that are 10 X values. So are 10 X. Values are going to be at zero first. So zero is just equal to zero times three is equal to zero. Our next one is gonna be at 10.3 point three times three would be 30.9. The next one is gonna be .6 .6 times three would be um 1.8 And then at .9.9 times three is 2.7. Um 1.2 times three is 3.6, 1.5 times three is 45. um and then 1.8 times three is 54 And 2.1 times three is 6.3. And then we have 24 times three, which would be 7 to. And then lastly we have 2.7 times three It should be 8.1. Okay so now we just need to add up all these values. So if we add up Um if we do that and we add up all the values and then multiply by .3 our area is going to be equal to 12.15. So we can see that as we um increase the amount of rectangles are. Area here is getting bigger and it's getting closer to what the actual area under the curve is.

We will find the approximate area under the curve described by the equation why equals one over skirt of X plus one between three and eight. By dividing the total 38 into five. Sub intervals in for a and tens of intervals in barbie and then add up the areas of the inscribed tanks. The height of your tangle may be found by evaluating the function for the proper value of X. So let's define the function F of X Equal one over square. It effects plus one. And we consider that function defined On the close interval 3 8. And we know that when x increases The fraction one over spirit of expose for decreases because one is a constant and the value in the numerator is greater. Greater And for a greater value in the numerator, two fractions smaller. So fc dressing on 38 and it's a positive function. So you can talk about area under the graph or the curve defined by dysfunction. Mm hmm. And we can see that we have something by this. Uh three. Let's see here The image is one of her squares of four which is 1/2. Let's see here. So we have this point on the graph and left and pine and eight. The image of F is one over squares of nine which is one third let's say around here. So this is the writing point of the graph function. Something like this N degrees in function. So for that reason we got to take or any Partition of the interval 3 8 to consider direct table inscribed under the area of uh the curve under the curve. We gotta take the writing the image of the writing point, the soup interval us the height of the routine. And that's for any sub interval. That's because the function is decreasing On the Interval 3 8. Okay. So that said we're now considering part eight and this is equal to five. So the step size H is eight minus 3/5 which is 5/5 evil one. And because of that notes are it's not equal three. I just want to go for X 25, three equals six four equal 7 and x eight X five sorry 48. Then we can say that the internal from 3 to 8 of one over Squares of excellence, one Is opportunity equal to age. There is one times The image of the writing points for the interval from 0 to X one. We got to take the image of X. Wanna see height Plus from X 1 to X two. We got to take the image at X two Plus from X 2-3. We've got to take the image at three From X- 32 X four. We got to take the image of four. And finally from X 42 X five. We've got to take the image that eggs five. And with this we had the some of the areas of the inscribed tangles for 5, 7 tables of the same size a Chingola. So this is F add poor This f. five. F 6 was F at seven plus F at eight. And we get one over Square to five. This one over squirts of six plus one over skirts of seven. This one over squares of eight plus one over squares of nine with his third one third. And I didn't up this a computer or calculator. We find that The intro from 3 to 8 of one over square root. If x plus one is then approximately equal to 1.92 0313 zero a two 899 655. And this is the approximation to the integral. Use him fives of intervals Of the same size age equal one. Okay. And uh the heights of each uh tangles equal to the image of the function at the right imports And that because the function On the interval 3 8 this situation. Okay. So now in part B we take an equal stand so we get ages Eight minutes 3/10, which is 5/10, which is 0.5 with this age we can say that the nodes are It's not equal three X one equals 3.5 X two is 4. X three is 4.5. X four is 5. X five is 55. Access is six At seven is 6 .5. x eight Is seven X 9 is 7.5. and extent 10 is eight. Well, so now we can say that the endorphins 3-8. 1 over square. It effects the swan separate. Similarly equal to 0.5 times. And then again we gotta take the rat the image of the writing point of each sub interval us. Um The height of the described tangle. And so that that lead us to the some from Ike was one because the function is increasing 2 10 of FX. Survive. And that is 0.5 times. Some from Ike was 1-10 of one over the square root of X. It by uh plus one. And that's equal to 0.5 times one over it started except one. So is 3.5 plus 44.5 square. It's a four point five. This one over square of five Less. one over square to five 0.5. This one over skirts of six sorry loss one over experts of six .5. This one over skirt of seven plus one over square it to seven 0.5 Plus one squared 8 Was one over squared off. Eight .5 Plus one over square tough nine. She's one third And isn't a calculator is approximately equal to one 95 92 for eight for 2431 67 zero three. That is centered from 3 to 8 of one over square root of X. This one separate simply equal to 19592484 24316703. You seem tens of intervals and inscribed tangles, which in this case are constructed taking as hide the image of the writer point of All the sub intervals because the function is decreasing on the interval 38. And we expect this approximation here to be more accurate or better than this previous approximation, in part because we have taken in part be more sub intervals.

We want to find the approximate area under the curve described by the equation Y equals one minus x squared between x equals 0.5 and x equal one By dividing the interval 0.51 into five sub interval first in part a. And then serve intervals in poverty. Then we had up the areas of the inscribed rectangles. The height of the church tangle might be found by integrating the function for the proper value of x. So would you find the function F of Mexico 1 -1 sq on the closed interval From 0.5 to one. So let's sketch this function there and we can rewrite this one man is X squared us one minus x times one plus X. And with this factory ization can say that the function Z equals zero at 91 1. In fact there's a problem open and what's because the coefficient of x squared is negative. So we can sketch the graph of this problem. This way We have woods near the one one and we know that the vertex or maximum value of the brow block course at the midpoint between the two routes that is zero. The value is one in fact. So this problem is like this more or less. We are interested only in the bar from 0.5 2 1. That is only is part here as we can see this function there is always increasing. So f is decreasing on 0.51 And for that reason and you're right to draw that little bit detail here, suppose we have something like this That we have a 0.5 here and one here. So when we take any partition of the interval to have the rectangles inscribed under the area Yeah. Which is comprised between the curve and the X axis between cerebral family one. We got to take the images of the writing points of this of intervals. That way we know that the rectangles will be inscribed, there's a curve and that for all points in the for any partitions. So we started by a we calculate the step size age And it's equal to 1 -0.5 over Members of intervals in part a which is five And we get 0.5 over five which is 0.1. So this is aged 0.1. And with this, who know that? The notes? All right, it's not equal. and 0.5 X one is 0.6 X to the 0.7 8 3. 0.864 0.9. And it's five is one. Then the interval from 0.5 to one of 1 -6 square will be approximated by the some of the area of direct angles. And we've got to take the images of the running point of the save interval studies Age 0.5, 0.1 times F at X one. That is the image of the running point of the Sioux interval from it's not too big swan plus F At X two. Which is the running point from the interval from next month to external and so on. If extreme plus F explore plus F&X five. And that 0.1 times The image of 7.6 Plus the image of 07. Lost the image of 0.8 Plus the image of 0.9 plus the one Which is 0.1 times 1 -0.6 sq plus one minus 0.7 square plus one minus 0.8 square plus one minus 0.9 square plus one minus one square. And these two terms cancel out because they are equal and This is then equal to 0.1 times we have 1234 -0.6 sq -0.7 sq -0.8 Square -0.9 sq Sorry, That's it. That's one. And we use a calculator and we found the disease. See your point 17 sold the interval from 05 to one oh oh 1 -6 sq. This opportunity legal. 2 0.17. When we use five of intervals. And the inscribe er tangles in this case. Taking for each of them as hide the image of the writing point. Now if barbie we use an equal stance of age is one -0.5 over 10. That is 0.5 over 10.30.05. With this age step size, we can now see what the notes are with them. We have eggs Not is 0.5 X one is 0.55 X two is 0.6 30 points 65 X 40.0.7 X five is 075 X six, X seven, 0.85 Eggs, 0.9 x 9 0.95. An extent is one. Then the interval from 0.5 to one of 1 -6 square will be Opportunity Lee equal to 0.05 times Some of the images from equals 1 to 10 of the notes. That is 0.05 times His son from I was 1 to 10 Of one -X by Square. And that is that's equal to 0.05 times 1 -11 square is 0.55 Square plus 1 -0.6 square plus 1 -0.65 sq Plus 1 -0.7 sq Plus 1 -0.75 sq plus one minus 0.8 square plus one minus your point 85 square plus 1 -9 sq See or nine, Sorry. And square Plus 1 0 for 95 sq And plus 1 -1 square is to cancel out. And then we get zero point See her five times 1, 56789 -0.55 sq -0.6 sq zero 65 square and a 0.7 sq -0.75 Square -0.8 sq -0.85 sq. And a 0.9 sq -0.95 sq With calculator. We found this at zero 18 nine 375. So the integral from 0.5 to one of 1 -6 square 17 is equal to 0.18937 went muse, uh tens of intervals and inscribed rectangles, which in this case corresponds to taking the hide secrets, the writing point, the images of the writing point of the seven triples. We all suspected this approximation to the interest is more accurate than the previous one because we took more sub intervals.

Okay, so we're looking at the function Y is equal to one over X squared. We're going from X equal to 12 X equal to five. And we're going to estimate it using rectangles this case four rectangles since N is equal to four. So adult to X is equal to our total width, which would be our upper bound, which is five minus are lower bound, which is one. So we have four divided by N. Which is four. So this is equal to one. So now we can say our area A is equal to one, multiplied by the Y values at the X. Values that we're looking at. So Y one Y two up to Y. N. And so the thing that we have to um be careful about is how are we going to pick these rectangles? So if we're looking at a graph of the function, why is equal to um one over X squared? It looks something like this. So You have this function that's going up to infinity as X goes to zero and then it goes down like this. And so if we said this was X is equal to one and this is X is equal to five. And we were going by ones in this case like this to actually um make rectangles under the curve. With the way that we're gonna have to do is we're gonna have to take right handed rectangles, so we're gonna start at X is equal to two. Like this, this will be our first rectangle and then our second one will be at X is equal to three And then x equal to four and then X equal to five. Since if we were to take left handed rectangles. So if we tried to take left handed rectangle, starting at X is equal to one, then we would be going above the curve at each of our rectangles and we don't want to do that. We want to be under the curve in um in this case so we don't want rectangles like this so we're gonna have to use right handed rectangles. So in that case we started X is equal to one which is our adult x value plus our lower bound. So our area I'll keep in red our area is going to be equal to one. So we can just get rid of that sentence, one multiplied by something and then our first X value is going to be delta X plus or lower bound which is one plus one just to so we have one divided by two squared which is 1/4 and then one divided by three squared which is 1/9 one divided by four squared which is 1/16 and then one divided by five squared which is one 25th. And if we add these all up in a calculator we'll see that this is equal to about .464. So this is this is our first area estimation using and is equal to four. Now we're going to look at it using N is equal to eight. So are adults X. In this case would be equal to our total width which was four divided by our end value which is eight. So this is equal to one half. So then our area here is going to be equal to one half, multiplied by the Y values at the X values. We're looking at we start at delta X plus our lower bound. So we're gonna start at 1.5 or three halves. And we're going to end at our upper bound since we're taking right handed rectangles. So 1/3 have squared is just two thirds square. We take the reciprocal of the denominator and two thirds squared is 4/9. And then we're looking at 1/2 squared which is 1/4. And now we're looking at um 2.5 1/2 and a half square to 1/5 have squared which is same as to have to fit squared which is for 25th. And then we have Looking at X. is equal to three now which is just 1/3 squared which is 1/9 and looking at 3.5 which is seven halfs which is same. So 1/7 half squared is the same as to seventh squared which is four 49th. And now we're looking at X. is equal to four. So we have 1/4 square which is 1/16 and 4.5 is nine halfs. So we take the reciprocal of that and square two nights squared is for 81st. And then lastly we're looking at X is equal to five, so we add 1/25. And now if we add all of these fractions up in a calculator and multiply by one half, we're going to get an area of about .5995. So we can see that again. Our area is increasing and it's getting closer and closer to the real area as we increase the the end value.


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