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Find the area ofthe region bounded by y to the right of the y-axisand y = 2 _ 2r" and3/ 15 6 / 17 44/15 40{17 None of the aboveSet but not sof-e the integral f...

Question

Find the area ofthe region bounded by y to the right of the y-axisand y = 2 _ 2r" and3/ 15 6 / 17 44/15 40{17 None of the aboveSet but not sof-e the integral for the a70 length of the cuve y =2 + 1 for <r < 2.Vi-erdrV1_4"dr2TV1+42dr2TV1+lr?dr None of the pove

Find the area ofthe region bounded by y to the right of the y-axis and y = 2 _ 2r" and 3/ 15 6 / 17 44/15 40{17 None of the above Set but not sof-e the integral for the a70 length of the cuve y =2 + 1 for <r < 2. Vi-erdr V1_4"dr 2TV1+42dr 2TV1+lr?dr None of the pove



Answers

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ 2 x-3 y=0, x+y=5, y=0 $$

The definite integral for this problem can be given by so I'm just writing the formula integration off minus 6 to 0 z off X minus fo fags the X going forward and solving it by putting the value minus 6 to 0 zero minus minus under root off 36 minus X square DX going forward and solving it further. I can write the value. Add integration off minus 6 to 0 Underwood 36 minus X squared dx. So solving it further I can write Devalue edge From the required given data, we can conclude that 1 ft off this circle has radius are equal to six years Now I am going toe calculate the area as we know that required idea id AEA, for this reason is equal to Pirie Square by food. So just putting the value of radius Year one by four multiplication by multiplication 60 square and on solving. I get 28.2743 as our final answer

Today we're going to solve program about 61. The required area bounded between G off X equals one and F of X equals seven. Divided by 16 minus X square. That is in the girl minus 3 to 3. Okay, Geo fix minus a 46 The it's integral minus 3 to 3 one minus seven. Divided by 16 minus X squared Big so began writing minus 3 to 3 one minus seven. Developed by 82 for minus six seven. Divided by 82 4 plus six The x x plus some anti IgE in tow Filling X minus four 78 Italy Express Full ministry to treat, which is six minus seven by eight, 11, 7 minus seven by eight Elements of it. We will get like 2.5947 Thank you.

The, uh, definite integral. The definite integral off this problem can be given by So I'm just writing the value 0 to 4 g off X minus f off x the x so going forward and just writing the formula again by putting the value zero minus minus under route 16 minus X squared DX now going forward and solving it further so I can write the value Age integration of zero to full under route 16 minus access square DX. From the graph from the grab data, I can conclude the radius radius off this radius off. This figure will be equal to food now, going forward and solving this so I can write the required area required area for this problem ad by our squired by food. So just putting the value off are here and on solving it. I get to 4.566 is squired units as our final answer

So in this question we want to find the area of the region enclosed by these curves. So we have this time two x minus three, Y equals zero. And then we have X plus Y equals five. And then finally we have y equals zero. So let's figure out what these lines all look like. Right. So I would put these first two guys into why equals form to begin. So the first one I've got negative three Y equals negative two X. Were equivalently y equals 2/3 x. This second guy I've got Y equals negative X plus five. So let's think about sketching. We've got Y equals two thirds X, roughly speaking, going to look something like this and then I've got Y equals negative X plus five. So that's a Y intercept of five Slope of -1 Y equals negative X plus five. And now I've got Y equals zero. That's just the X axis. So that guys down. And so my region of integration I believe is right in here this time. Now I could set this up in the dy dx direction but that will require too many girls. I think it's easier to set this up. Dx dy But before that let's figure out where these guys intersect. So these guys intersect when 2/3 x equals negative X plus fall. If I add X to each side, I'm getting five thirds X equals five. That's giving us an x coordinate of three. That's pretty nice. My ex is three and my why is two for this intersection. OK now I'm going to do this with a single dx dy double interval. Um integrating the function one. Now I'm going to do this in the dx dy direction. So let's see if I have a line parallel the X. Axis. Where would that line enter My region? You would enter when X equals three halves? Y. So that left boundary is X equals three heads. Why? And we would exit when X. Is what? Well Y -5 equals negative X. X equals five -Y. So I'm exiting my region When x equals five mines. Why? So I'm going from X equals three halves wide. X equals five minus Y. As my wise range from 0 to 2. As my wise go from zero to we said the y coordinate of intersection was shit. And so now let's see what happens. I have the interval from 0-2. My anti derivative of one with respect to X. It's just X. Being evaluated from 3/2. Y 25 -Y. De waal that's giving me The integral from 0 to 2. I plug in my top limit of integration five minus why minus what I get when I plug in my bottom limit of integration three halves. Why dy combining like terms I've got negative y minus three halves. Why? That's negative five has Y. So my integrate here is 5 -5/2. Matted to evaluate I'll need an anti derivative. My anti derivative of five is five. Why? My anti derivative of five halves? Why? With respect to Why is 5/4 wide square, Evaluate 0- two. We're getting 10 five quarters times four. 10 -5 is five. I am getting five as my final answer for this question. Hopefully, that makes sense. Have a great rest of the day.


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