Score: 0 of50 of 100 (100 complete}HW Score: 94.5%, 94.5 of 1005.6.107Question HelpFind the area of the region in the first quadrant bounded on the left by the y-ax...

Set up a definite integral that represents the area bounded by the graphs of the indicated equations over the given interval. Find the areas to three decimal places. $$ y=-\sqrt{100-x^{2}} ; y=\sqrt{100-x^{2}} ;-10 \leq x \leq 10 $$

The area bounded by two curves on a closed interval from a to B, um is determined by while the area A is equal to the integral, going from a to B off well, f of y minus g of Y D Y, where we divide these shaded region into two parts. Column A one and a two. And then we evaluate, um, each of them separately and then add them up. So for the limits of integration of a one, we have three minus y is equal to two times the square root of why So we just square here and this implies that three minus why squared is then equal to four wine. We just square both sides here to the square root here, um, and then, well, three minus y squared. Um, this is gonna be, what, nine plus y squared minus six. Y, um is equal to four. Why, we said is equal to zero, and we have y squared minus 10 Y plus nine is equal to zero. This is gonna factor as why minus one times Y minus nine is equal to zero. Okay, so now we have, um Well, here we just take y 01 because Y is equal to nine. Um, doesn't satisfy here, So we take y being equal toe one. Okay. And here we have a equal 01 and b equals one. And we have f of y is equal to two times the square of y and g of y is equal to zero. Therefore, we have a sub one is equal to the integral, going from zero toe, one off, two times the square root of Why de y Okay, if we evaluate this, um, this is gonna be equal to Well, we have to y to the three halves divided by three halves, evaluating from 0 to 1. So we get four thirds and then just times while one minus zero. So this is equal to four thirds. And then for the limits of integration of a sub to I'm second area here. Well, we have why minus one squared is equal. Thio three minus. Why? So that implies that why squared? Um, well minus two y plus one is equal to three minus y. So you confess is equal to zero and we have y squared minus Y minus two is equal to zero So we factor this. This implies that we have why, minus two times why plus one equal to zero. And here we take that y is equal to two because you have y being greater to zero. So one doesn't satisfy here. Okay, so again we take a equal to one, be equal to two, and f y is equal to three minus y and g f y equal to y minus one squared. And that therefore we have a sub to equal to the integral, going from 1 to 2 off well of three minus y minus. Why? Minus one? Whoops. Why? Minus one quantity squared de y. Okay, so we evaluate this and we get this is equal to get the integral from 1 to 2 of just, uh, two plus. Why minus y squared de y. So therefore, this is going to be equal to while we get two y plus y squared over two minus. Why so on stretches down here. We get to why. Plus why squared over two minus y cube over three. And then we are evaluating this from 1 to 2. Okay, so we first plug in to instruct off. We give you plug in one. So we get four plus two minus a thirds and then minus to plus one half minus one third. So that gives us four minus one half minus seven thirds, which gives us 21 minus 14/6. This is equal to 76 Okay, so therefore, the total shaded area is just equal to a one plus a two. That's just equal to four thirds plus 7/6. So that's just equal to come. Denominator. Here it's gonna be equal to eight plus 7/6, which is equal to 15/6. So therefore, um yes, the solution here, the area bounded by these curves is 15 6. All right, take care.

So we want to find the area of this region that's pictured here. So we're given several functions of X in terms of why, and you see how maybe we can sort of separate this into two different regions that are bounded by two curves. So if I draw this horizontal line, X is equal to sorry. This horizontal line Y is equal to one we see below the line. The region is bounded to the left by just the Y axis that's X equals zero and to the right, by this curb, excess to square roots of why and above that our upper bound is three minus wise, equal X, and are lower bound his exes. Why minus one squared. So the area we can write down as being the integral first from 0 to 1 with respect to why of to square roots of why minus zero. Okay, because the Y axis is the graph of X equals zero. All right, so then yes, sir, than d y here. And then we would just have the integral from 1 to 2. Now what's our upper function is three minus y and then our lower function is this one red. Okay, so these two in a girl's add that together gives you the total area of the shaded regions. Now we won't evaluate thes because it should be straightforward. You just expand here on the right hand side. You're just integrating polynomial. So to find the anti derivatives, just reverse the power rule. If you're confused about how to actually integrate thes these two look back to some previous videos where we work through all the steps and full detail.

Mhm. I have given that these inequalities here. We're just profit here. The problems that we have Y equals X plus one. The spread line of Debian wife for the last one than we have Mexico's truth. But we have the results communicated by giving half the street and the originates in april The required areas. You wanna, we have a three well people This point domestic 1- one. Poetry is we have to go to one of those data. Little one the graph that we region under this Parabola here here. Black one, the X blah 1-2. We have this region here. one or 2 That is given an under explored one. Yeah. So it is. So this now so you get executed over three Protect from 0 to 1 last expert over to. Well, thanks from 1- two. It's hard to be happier. 1/3. Last one Plus we have uh two plus 2 -1/2 -1 1, 2000. You have to be heading up or negative one over to the 1/3 that's coming out to be negative. We have three positive too. So that's coming on to be 23/6 after required area. We'll say goodbye. Although the region between this and the by not in first warden all tv have exposed to in the first quarter and begin exports to and we have actually Y plus one. Uh this problem. So next to go to Mexico too heavy in the first quarter than this Media and the next critical one. Every once we get this region here, that is there too. And then we have this vision here that is given as anyone find out area when a native way one would be this negative. This, wow, it was this negative this problem maybe just why not done To have a one? So they probably have 0-1 X plus one negative X squared plus one. Or be it. It's called this. You'll get 1/6. Then you too. It's coming up that we wanted to we have expert plus one negative X plus one more the X. That gives five Olympics. The required areas. They want potato. They're coming after one of these blood piper will take that as being one where you know, I'm supposed to be partners once per unit and we have done so for the both parts that has given us 40. You always look at the graph is already good. Thank you.

We want to find the area bound by the curves five X minus nine and X squared minus two X plus one that is this a little sliver in here? So first, let's figure out where these two curves cross each other here and here. These will be our bounds for the integral. So it's set five X minus nine equal to X squared minus two x 1st 1 So if we move everything to one side, we'll get X squared. Nine x seven X plus 10 equals zero, and if we factor, we get X minus phi times X minus two so our curves cross each other at X equal two and X equals five. Now let's set up are integral and grading from 2 to 5, and our top curve is five X minus nine, and the bottom one is X squared minus two X plus one. So if we just simplify this a little bit, we'll get negative X squared plus seven x minus 10 which integrates to negative 1/3 x cubed plus seven haps X squared minus 10 x from 2 to 5. And if we plug in five, we get negative. 25/6 minus what we get if we plug in to which is negative. 26 over three. So if we find a common denominator here we have negative 25/6 plus 50 to over six, which gives us 27/6 or nine, Hafs. And this is our calculated area of the bounded region.