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Find the area of the shaded region by (a) integrating withrespect to x and (b) integrating with respect to y.(Graph cant copy)

$$9$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

Integrals

Integration

Applications of Integration

Area Between Curves

Volume

Arc Length and Surface Area

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So here we have. Why equals X squared minus two and y equals X, And we want to find the area of the shaded region. But we've been asked to just set it up. We're just going to set up the integral we're gonna set it up for both. With respect to X and with respect alive, we don't actually have to solve the integral, which is gonna save us a lot of time, actually. So first, let's look at this. We know that our boundaries are the X axis, the line y equals X in our parabola X squared minus two. And we know we're gonna need these three points to help set our boundaries for any girls. So let's go about finding those points first. We know this point is 00 because y equals X passes through the origin. That was easy enough. We're gonna need this interception point where y equals X squared minus two passes the x axis. So if we set y equals zero there we get zero equals X squared minus two, which factors? Two X plus radical too and X minus radical, too. So we have the excess intercept at X equals plus or minus groups, plus or minus radical, too. Now, our point is in the negatives we're not We're not worried about this point here in the positives. So we care about our negative. So the point we're gonna use is appointed negative, too. So this point here, his negative radical, too zero. And then we need this point here, which is our intercept point. So intercept is where they're equal. So we get X equals X squared minus two or zero equals X squared minus X minus two. And this factors two X minus two and X plus one. So we have intercepts at X equals two and X equals negative one. Now, again, we care about our negative x value because that's what Quadrant Warren. So we're not concerned with our positive value were concerned with our negative value. So we care X, isn't it minus one? If we plug X is minus one into y equals X, we get negative one and one. So that's our final point here. Negative one negative one. So we have our three points where you're gonna represent our boundaries, and we have her functions. So let's look first with respect to X We're gonna set up an integral we have A equals. The area equals three integral from A to B of F of X, minus G of X. And if we consider, are vertical line test here. We're gonna need to different inter girls and they're gonna be split here at negative one. So let's set those up. We have the integral from negative radical, too to negative one. And our top function is just the X axis. So it would be zero and are g of X are bottom function is gonna be y equals X squared minus two. So he would actually have zero minus X squared, minus two. So we're just gonna put a negative sign out front. We have negative X squared minus two DX plus our second integral, which is gonna go from negative one 20 And here, where y equals X is our G of X. And again, our f of X is simply zero. So are integral is going to be negative. X dx. We can simplify that a little bit. There's very little change, but we'll do it just for your simplicity's sake. It's the integral negative radical to negative one of two minus X squared DX minus the integral from negative 120 of X DX. Both are correct. Both are set up on. That's all we need to do. Now. If we were to do this with respect to why our area is integral for May to be of effort, why minus g y de y, uh, And if we look so what we're gonna need to do here is we're gonna need Teoh get our equations with respect. Why so y equals X simply becomes X equals y and y equals X squared minus two is going to become X equals plus or minus radical y plus two. Now again, we are in the negative quadrants where X is negative and in fact, the region wearing his exes and wire negative. So in this case, we care about negative radical y plus two because this is going to represent this half of the parabola that's on the left side of the the Y axis. So that's what we're gonna use. We have X equals Y and X equals negative radical Y plus two. Now, if we look here and we do the horizontal line test for with respect to y we noticed we only need one Integral because we're going to hit the same two functions all across our boundary. In this case, our boundaries gonna go from negative one 20 So are integral is from negative. 120 r f of x is our right most equation, which is why equals X. It's our line. Or, in this case, X equals y Why minus negative? Radical? Why plus two. Do you, I or Teoh simplify the negative signs? We have the integral from negative 120 of why plus radical y plus two de y and that's it. We don't actually need to solve these into girls, but we've set them up.

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