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Find the area of the region I, and = 2in the Ty-plane bounded by the graphs: %and...

Question

Find the area of the region I, and = 2in the Ty-plane bounded by the graphs: %and

Find the area of the region I, and = 2 in the Ty-plane bounded by the graphs: % and



Answers

Find the area of the region bounded by the graphs of the given functions. $$ x=-y, x=2-y^{2} $$

In Brooklyn. 64. We want to get the area bounded by the grips. Why equals Eat the board of X and y equals Eat the board of Negative X and X equals minus two. Let's scratch these graphs to visualize about the dream. We have your ex. We have your wife. Let's sketch. Why equals into the world of X when X equals zero? Why equals warm? Then we have here a point one four. Why equals you to the bar of X when X equals one when X equals one y equals e which is approximately 2.7 here too. Yeah, we're three then. It's something like here because it's exponential. Then we never go below the X axis. And it goes something like that when X it's an exponential. And why is never Zoo goes here toe negative infinity. The execs You goes here up. This is for why equals use of r of X. It's a graph into the bar of minus X. It will be the same but similar when X equals zero. We have the same point here when X equals minus one. We have the point E Why equals e which comes in black here and it's exponential. We never goes below zero and it goes something like that. Tenders to infinity here, positive infinity and goes up here. The line X equals minus two is a vertical line at X equals minus two. This line, we want to get this the area off this region. After a visualization of the area, we can see that this area is bounded at the top by the function. Why equals e to the part of minus X, and it's bounded in the bottle by the function. It is a bar of X. Then we can use definite integral together the area. The area equals the definite integral off the top function. It is a part of minus X minus the bottom function the X and we we integrate for the bounds off the region. The region starts at X equals minus two, which is destroyed. And in this at the intersection off the two graphs at X equals zero from minus two 20 Let's integrate the integration off it on the board of minus X is eat a bar of minus X and we divide by the differentiation off this polynomial we divide by minus one minus. The integration off it is a lot of ex is It was a lot of X and we integrate from minus 2 to 0. First we substituted by the upper bound. We suggest uber X equals zero. We have it with a bar of zero. She's one divided by minus one minus one into the world. Zero gives one which is minus two minus one minus one. She's minus two minus. We substitute boy, the lower bound be subsumed by X equals minus two. We have minus E to the bar off minus minus two, which is two minus. He is a part of minus two equals. We have your minus two and we have here minus minus. He to the bar off to minus E to the about off minus two. Give us minus 7.5 to 4 and we have here a minus. Sign 7.5 to 4 and we subtract two from it in the gifts 5.5 to 4, which is the area off the bounded Asian. Well, you, the graphs given in the problem and the final answer off our problem

As you look at this problem with the curves, Y equals X. It's just a diagnosed line and then why equals they have is two minus X. To the Y intercept is too and then it's down one right one. And then the third equation is Y equals zero, which is the X. Axis, which is kind of nice because it just tells you, hey, the area we're looking for is this little triangle that has a base of a width of two. So if you think of the area of a triangle As 1/2 times base times height and you start to fill it out and like you know the basis to And the height they intersect at one. The answer half of two is 1 times one is 1. I'm not sure if your teacher would let you get that answer. But the actual faster way would be to take this blue equation and rewrite it as adding X over and subtracting why to the right side and then doing the problem In terms of why? So what you can do is the integral from 0 to 1. The y values The upper function is that two -Y. And the lower function is just X equals Y. So subtract off why? As I mentioned in terms of why? And as you simplify that B0-1 of the function of two -1 -Y is -2. Y. Do you want? And you can find the anti derivative of that Adding 1 to the expo and divide by your new exponent From 0-1. And as you plug in your bounce Uh two times 1 is two -1 Squared is one. And I'll go ahead and put in zero as well. Although it feels like a waste of time because subtracting zero is not going to change the value of anything and 2 -1 is one. So we have the correct answer. As predicted Answers. one The area is one

Embroiling 23. We want to calculate the area Enclosed boy. This line and this graph Let's sketch of them if we have your ex envoy One, 23 I have 123 That's through the line. We have y equals X plus two. It's when X equals zero. You know that. Why will be to you Have a point here and the slope is one. This means we have here a point. And here a point and so on. And you're a point. This is the line. Why equals explodes too. The second graph is y equals X Quit is something like that. We have here one on one and at minus one. We have a buoyant here and we have zero and zero Those when X equals two. We have y equals four here. This point is shared between the two functions. Then we have a point here. The point here point here. A point here and so on. We have this graph is boy equals x squared and the area in between the area enclosed is this area. We can get this area using the definition of different integral the integration off the top graph the equation off the top graph minus the equation. Off the bottom graph, we have the equation off. The top graph is why equals X plus two minus the equation off. The bottom graph is y equals X squared. We have DX here, and the bones is the starting point of the bounded area here and the starting point of the bounded area. Here we have the bounce from minus 1 to 2 to make sure that the two points intersects at X equals minus one and X equals two. We can substitute here by X equals minus one. It gives why equals minus one plus two equals one. And when we substitute by X here equals minus one, it gives why equals one. This means they intersect at X equals minus one. To make sure the intercept at X equals two. Substitute by X equals two in the line to two plus two gives four. And when substitute in the function here by X equals two, we get to power. Two equals four and this means they intersect. They gave us the same point, minus one on one, two and four, meaning that thes bones are through our correct Let's integrate Integration of X is X squared, divided by two. Integration of two is two x minus. The creation of X squared is execute. Divided by three we integrate from minus 1 to 2. We started by the upper bound we substitute by X equals two to the bar of two, divided by two plus to multiply it by two minus two cube divided by three minus. We substitute by the lower bound X squared divided by two equals half minus xmas to block by minus one is to plus one third. Let's evaluate these two terms. The first term is to plus four equal six minus eight, divided by three equals 13th minus half minus to plus one third equals minus and we have another minus 7 12 by six equals nine, divided by 24.5, which is area off the shaded region and enclosed by the line and the curve, which is a final answer off. Our problem

We are asked to find the area founded by these two curves. So the first thing we should do is sketch this out should be a very rough sketch and if we do that and then we left with something like this. So we're curious about what is, what is the area between these two regions. So in order to figure out our bounds which can be right here, it's gonna be at the intersection point, you're gonna have to set these two equal to each other. So you can say that two divided by one plus x squared Is equal to one. If we multiply the denominator over We find that too is equal to one plus X squared. Now we can subtract over that one such that we're left with one is equal to X squared. Then we find that X is equal to negative one positive. So this can be our bounds of integration here and we find area between two curves. We take the upper curve and subtracted with the lower curve, subtracting the lower curve from the upper curve. So as we can see here from our graph that this is our upper curve which corresponds to why is equal to two divided by one plus x squared. So we can set this up as the integral From -1 positive. one of our upper curve minus our lower curve. This will be integrated with respect to X. We integrate this our first term in the instagram here too divided by one plus x squared. This becomes two times the inverse tangent of x minus one. Well integrate up to minus X. So we evaluated at one negative form. So we can plug into evaluate now this will be equal to two times the inverse tangent of one. Make that a little more clear minus one. We're going to subtract two times the inverse tangent of negative one -12. So it becomes plus more that we can distribute in the negative here. So we'll write that out. So we're left with two times the inverse tangent of one minus one -2 times the inverse tangent of negative one minus one. So we can combine these terms here, Which leaves us with two times the inverse tangent of one -2 times the inverse tangent of negative one minus two. And we can leave that as our final answer. But let's simplify it a little bit. This is the same as Pi -2.


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