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Current time:0:00Total duration:8:03

AP.CALC:

CHA‑5 (EU)

, CHA‑5.A (LO)

, CHA‑5.A.2 (EK)

so I have two curves graphed here and we're used to seeing things where Y is a function of X but here we have X as a function of Y in fact we can write this top expression as being a function of Y and the second one just to make it different we could view this as G of Y once again it is a function of Y and what we're concerning ourselves with in this video is how do we find this area in this light blue color between these two curves and I encourage you to pause the video and try to work through it alright so a huge hint here is we're going to want to integrate with respect to Y a definite integral where our bounds are in terms of Y so for example this is this lower point of intersection right over here this would be our lower bound in terms of Y let's call that y1 and then this up here this would be our upper Y bound so if we think about where do these two curves intersect and we look at the Y coordinates of those intersections well that gives us two nice bounds for our integral so we're gonna take our integral from y1 to y2 from y1 to y2 y2 we're going to integrate with respect to Y dy and so what are we going to sum up well when we integrate we can think about taking the sum of infinitely thin rectangles and in this case it would be infinitely flat rectangle since we're thinking about D Y so dy would be the height of each of these rectangles and what would be in this case the width or the length of this rectangle right over here well over this interval from y1 to y2 our blue function f of Y it takes on larger x-values than G of Y so this length right over here this would be F of Y F of Y this x value minus this x value minus G of Y so this is going to be f of Y - G of Y G of Y well we know what F of Y and G of Y are really the trickiest part is is figuring out these points of intersection so let's think about where these two curves intersect they are both equal to X so we can set these two Y expressions equal to each other so we know that negative let me do it in that other color so we know that negative y squared plus 3y plus 11 is going to be equal to this is going to be equal to y squared plus y minus 1 so let's just subtract all of this from both sides so then on the right side we have a zero and on the left side we just have a quadratic so let's subtract Y squared let's subtract Y and then subtract negative 1 which is just adding one is and then over here we're going to do the same thing - y plus one and what we are left with is going to be as hopefully a straightforward quadratic so let's see this is going to be negative 2y squared plus two why am i doing that right yeah plus 2y plus 12 is equal to zero and then this over here I can factor out a negative two and I get negative two times y squared minus y minus six is equal to zero this we can factor from inspection what two numbers when we add equal negative 1 when we take the product we get negative 6 well that would be negative 3 and 2 so this is going to be negative 2 times y minus 3 times y plus 2 this is straightforward factoring a polynomial or quadratic do I do that right yep that looks right is equal to 0 so what are the points of intersection the points of intersection are going to be Y is equal to 3 and Y is equal to negative 2 so this right over here is y is equal to negative two and then the upper bound is y is equal to three so now we just have to evaluate this from negative two all the way until three so let's do that I'm going to clear this out so I get a little bit of real estate so this is equal to the integral from negative two to three of negative Y squared plus 3y plus eleven minus all of this stuff so if we just distribute a negative sign here it's minus y squared minus y plus one and then we have a dy D Y this is equal to the definite integral from negative 2 to positive 3 of let's see negative y squared minus y squared negative 2y squared and then 3y minus y is going to be plus 2y and then 11 plus 1 plus well we saw this just now we were trying to solve for y dy and so what is that going to be equal to well we just take the antiderivative here this is going to be let's see negative 2 let's increment the exponent Y does they're divided by that exponent reverse power rule plus 2 y squared divided by 2 which is just Y squared just the reverse power rule and then plus 12y and we're going to evaluate that at 3 and at negative 2 so if we evaluate that at 3 we are going to get 2 negative 2 times 27 over 3 plus 9 plus 36 and then we are going to want to subtract minus all of this evaluated negative 2 so it's going to be negative 2 times negative 8 over 3 plus for minus 24 so we just had a little bit of mathematics ahead of us so let's see this is going to be 27 divided by 3 is 9 so this is negative 18 negative 18 plus 9 is going to be negative 9 plus 36 all of that is going to be equal to so the stuff in blue is equal to 27 all right did I do that right yet negative 18 plus 9 yep 27 and then all the stuff in red over here we have this is going to be negative times a negative so it's 16 over 3 plus 4 minus 24 so that is going to be 16 over 3 and then minus minus 20 minus 20 but then we have this negative out here so it's if we distribute that we'll get plus 20 minus we could say instead of 16 over 3 we could rewrite that as 5 and 1/3 minus 5 and 1/3 and so what is that going to get us let me scroll down a little bit let me go to the right so I have a little bit more real estate here so then that is going to be equal to you get a minor drumroll here 47 minus 5 minus 1/3 which is equal to let's see 47 minus 5 that's 42 minus 1/3 is now we get a drumroll 41 and 2/3 and we are done

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