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Find the area of the region bounded by the graphs of the following equations6r + &1=3 and X=7Hnnuinalls...

Question

Find the area of the region bounded by the graphs of the following equations6r + &1=3 and X=7Hnnuinalls

Find the area of the region bounded by the graphs of the following equations 6r + & 1=3 and X=7 Hnnuinalls



Answers

Area Find the area of the region bounded by the graphs of $y=7 /\left(16-x^{2}\right)$ and $y=1$

Sort of looking at this, trying to find the these two equations are trying to find the area between them. So let's go ahead and sketch of this here. Sweet got this region right there. Where we have this X 2016 here, that's when y zero and then we have X equal seven. Somewhere around there. And the region that we're looking at, it's between them here. So that's this area right here that I've shaded in blue. So let's go ahead and see where they intersect. We have 16 minus y squared equals seven. That means why squared is equal to nine. When we get Divide by also try 16 and divide by -1. Therefore, when we square root we have y is equal to plus or -3. We're going to be our boundaries. We have the integration from -323, 16 minus Y. Just on top might have seven. Dy That simplifies to nine -Y. Dy From -3 2, 3. And so that we have here, that's nine y minus why squared over to. And so then that's going to go from -323. Thing about this is that because it's symmetry to the X axis here and we're doing this with respect to why this would actually be the same thing as doing nine y minus Y squared over to two. If we do this from 0 to 3. And so then we do three times nine times three minus three squared over two, That all that Times two. So so That gives us 27 -9 over to. It's too multiplying the two and there we get 54 minus nine, So that's going to be equivalent to 45. That's our area.

Let's talk about questions. 60. We need to find the area of the score bounded by white. Below zero means X axis X equals zero, which is why access and execute, too. So let's start first. We're fact rising the denominator, so we'll split the middle term. We have X Square plus five x plus two weeks. In fact, not five weeks. And to accept will be four x and three x because four times three is 12 for express three x plus 12. So from here we're getting a state X out from here and three out from here. So we're getting X Times X plus four. Those three times Experts four. Let's take express four out again. So we have express four out on expressed three out. This is why. All right, so we can clearly see that Venezuela Positive survivors positive and the denominators positive. So here we have minus four. Here we have minus three. This is the positive region and this is the negative region on. We are interested between zero and toe clearly between zero into its positive because for all the values better than minus streets positive now let's find its value at X equals zero. What is its value at X equals zero will be 15/12. 15/12 is nothing but, uh, 5/4 and at X equal to its value will be 15/2 square, plus seven times two is 14 plus 12. This will be equal to 15 over. There should be 26 plus 4 30. So this is nothing but one or two Me, right? So we have the values as, uh, we have both the values at both the X. So let's find out this area for the area we know that X equals zero is somewhere here one point something, Let's say 5/4 on X equals one x equal to is somewhere here it's one or two. So the area will just is the graph connects like this Then the area is the and this is like sequel toe to the end in the areas of the shaded part. So for the shaded region, we gotta integrate the function which is 15 over X plus four Times Express three between zero and fish. Let's do that. Fifteens taken outside the numerator has one and one candidate and as X plus four minus X plus three, which is nothing but one over X plus four Times Express three. Another denominators. Common shit should be given toe both the new merit Er's So we have one of our express three here minus 1/4 Express here on here. We have DX. This integral is pretty straightforward. So we have both the integral Xas extreme. The integral sign. So we have forced integral as natural law of express three minus natural. Log off express four between zero and two. So this comes out as 15 times, uh, can very natural log off express tree over express for using the properties off. Lock between the limits of zero toe Do So this comes out as natural. Log off if you place too and we have five or six. If you play zero, we have three over four. So this is its answer. And in fact, we can further simplified. It has a log off a minus log B. There's nothing but log off. You will be so here. Will have Okay, it over be. This can be further return us log. You will Lobby will be five times 4/3 times six. So its final answer comm sorters natural. Log off 20/18 which is nothing. But then overnight. So this is Yeah, the final answer.

In probably 37. We want to get the area enclosed by these two parabolas. Its first sketch, These two problems. To visualize the shaded area, we substituted by different values off X to get boy for the first problem when X equals zero, we have y equals five x equals one we have war equals tu minus six minus four plus five equals one When why equals two. We have to multiply it by four She's eight minus 12 minus four plus five equals one. We have zero and five zero and five one on one two on one. This is a parabola for the first girl. Something like that. We don't know it goes below the X axis or not. Never mind. Let's go. The second parabola when X equals zero we have one equals minus 15. When X equals one we have X, we have y equals one plus 67 minus eight. When X equals two. We have boy equals for lost 12 equals 16 minus 15 Swarm When X equals three we have oy equals lying plus 18 violence 15 You have all equals 12. Let's start with the first point for the 2nd 0.1 on minus eight on I minus. It's something like here. 211 here. Three and the 12. Something like here. Then we have the problem. This problem because something like that And if we continue this problem, they will intersect at some point. That's a good the ex corded. Off this point, we can get this coordinated by equating the Y Coordinate off both graphs. We have to X squared minus six X plus five equals X squared plus six x minus 15. By solving this equation, we get two points. This is the first point we know it. It's X accorded equals two, and this is the second point. Let's get it's extraordinary. We subtract X squared from both sides. X squared minus x x plus five. Equal six X minus 15 We subtract six x from both sides. We have X squared minus 12 x last five equals minus 15. We add 15 to both sides, then X squared minus well X plus 20 equals zero. You want to get two numbers. Their multiplication is most of 20 and there some is minus 12. The two numbers is minus three and minus four. And there there's some is his minus 12. We have 20 can be to beytin. The two numbers is two and 10 x minus. Stoop The two numbers the minus two and minus 10 x minus two. Want to blow by X minus 10 equals zero. The first point is at X equals two, and the second point is at X equal. Stine. They meet at X Equal Stone this point as accorded equal stone. Never mind about the sketch, and it's a scale. It's not to skill. Let's use the definition off the definite integral To get this area. The area equals the top function a definite integral off the to function, minus the bottom function. That function is the second parabola here, and the Balkan function is the first bar apology. Then we have the definite integral. Off the second problem. X squared the blast X X minus 15 minus the first problem two X squared minus six X plus five The X We integrate for the bones off the area. It starts here at X equals two and ends at X Equal Stone or the solution off this equation. We integrate from 10 from 2 to 10 from 2 to 10 that's integrate X squared minus two X squared is minus X squared plus six x plus six x plus 12 x minus 15 minus five is minus 20. The X from 2 to 10 equals minus execute, divided by three plus 12 x squared, Divided by two When I was 20 x, we integrate from 2 to 10. We started by substituting by the upper bound with service Uber X equals 10 we have minus 10 pube by three plus six multiplied by then squared minus 20. Multiply by 10 minus. We substitute by the lower bound minus toe cube by the by three plus six. Multiply by two square minus 20. Multiply by two Evaluating the first term. Gives minus 1000 by by three plus 600 minus 200 equals 205 by three We have minus. Second term is minus eight, divided by three plus six months by before minus 40 equals minus, and we have a minus sign. 56 divided by three equals 256 Divided by three This is the area bounded by the two given problem in the problem

So we're working with the curve y equals seven over X squared, and we're going to be evaluating the area of the region under this curve. But it's bound by the X axis, so X, so why has to be positive and is also bound by X equals one. So we're going to It's found on the left by X equals one. So let's just write that in that That's the left. So if we were to draw a picture of this would see that we have this sort of curve going like that that goes all the way to infinity getting closer and closer to some number. Um and so we're going to be cut off by the X axis and on the left by Y equals one. So we're cut off right about here or by X equals one. So we're going to go from X equals one all the way to infinity. So when we're writing an integral, we're going to have one to infinity of our function. Seven over X squared, evaluated with respect to X. Now we can't go all the way to infinity, so we're just going to have to approximately a limit that as T approaches infinity. So we're gonna try to get as close to infinity as possible without actually trying to reach infinity. So now we're going to be able to take the anti derivative. So I'm going to rewrite this as the limit as T approaches infinity of the integral from one to t of seven times X to the negative too. So the anti derivative is going to be equal to the limit as T approaches infinity of seven times X rays, the exponents by one to the negative one. And now multiply our divide by the new exponents. So that's times negative one evaluated from one to t. So now we're going to plug in tea and one for our upper and lower bounds so we'll have the limit as t approaches Infinity of negative seven over T minus negative 7/1. So for this first term, as T gets larger and larger and larger in the denominator, the larger the denominator a fraction of this fraction the closer, the fractions, getting to zero. So we're just going to end up with zero when t is equal to basically infinity, and that's going to be added to seven. So our solution for the area is going to be equal to seven. Under this curve from X equals 12 X equals of


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