Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Area between curves. If G prime is continuous on the closed and bounded interval AB and F is continuous on the range of G of X equal U, then the integral AB of F of G of X * g prime of XDX is really equal to G of A to G of BFUDU. The proof if we start with the integral ABF of GXG prime XDX, that's really equal to capital F of G of X sticking in B&A for our X. So F of GB minus F of GA. Now if we come up with some new function U and we say let's let U equal G of X. So then we would have F of U from GA to GB. And if we took the derivative of that, we'd then have an integral of GA to GB of FUDU. Let F be continuous on the symmetric interval negative A to A. If F is even, the negative A to A of F of XDX is really just two zero to AF of XDX. So if we thought about this being even, we can see that the area under the curve from negative A to zero would have to be the same as the area under the curve from zero to A. So if we found just zero to A and doubled it, that would give us the area under the curve for an even function. If F is odd, the negative ETA a of F of X DX is going to equal 0 because the area below the curve and the area above the curve are going to be equal and they're going to cancel each other out. So negative ETA a of F of X DX is just zero if F is odd area between 2 curves. If F&G are continuous with F of X greater than or equal to G of X throughout AB, then the curve of the region between the curves Y equal F of X&Y equal G of X from A to B is the integral of F -, g from A to B or the area equal the integral A to BF of X -, g of X DX. Also a = C to D of FY minus GYDY. So what this is saying is that we can actually integrate doing vertical or we can integrate doing horizontal. Let's look at some examples. Our first example, if we're going from -1 to one of five r / 4 + r ^2 ^2 doctor, we're going to start by letting U equal that 4 + r ^2 in the denominator, because then DU is just two R doctor. Now there are lots of ways to do this, but frequently we get the variable that is in the original integration by itself. So if we just or divide it each side by two, we'd get 1/2 DU equaling R doctor so that when we do the integration, this RDR here is going to get taken away and replaced with 1/2 du. Now we had a 5 on the top and we know that constants can just come in and out of the integral. So I'm going to pull the five out in front and we're going to get one over instead of 4 + r ^2, we let that be U. So over U ^2. Now we can actually change the bounds. So if U equal 4 + -1 ^2, so U is going to equal 4 + 1 or five for the new bottom bound. And if we let U equal 4 + 1 ^2 for the top, we're going to get 4 + 1 or five to the top. Now if we look at this by our fact that if we have a bottom bound and a top bound that are the same, we know that this is really just going to equal 0. Let's say that we didn't see that, and we'll continue to solve this. So if we thought about this as being 5 halves from 5:00 to 5:00 of U to the -2 to U, add 1 to the exponent and divide by the new exponent. When we divide by a negative, it's just the negative coefficient. Now we can see when we put in the upper bound minus the lower bound, we're really going to get 0. Now a different way or a different? Let's just change the bounds on this. Let's instead of going from -1 to one, let's do a new problem using the same idea but going from zero to one this time. So we're going to start this the same way with the same substitution. So we're going to end up with the 4 + r ^2 being our U2. RDR is du, RDR is 1/2 du. So when we look at this, we're going to pull out the five and the half, and we're going to get 1 / U ^2 du. But now our bounds, now our top bound, because the one is the same, is still going to be 5. But when we put in AU of AR of 0 for the bottom bound, we're going to get U equaling 4. So now we're going from 4:00 to 5:00. If we do the integration here, we're going to get -5 halves, you'd and the -1. So we're going to get -5 halves of 1/5 -, 1/4. So when we look at this, we get 4 -, 5 / 20. So -5 halves -4 twentieths -5 twentieths, oops, +4 twentieths -5 twentieths negative, 120th negative times negative is going to make it a positive. The five and the 20 are going to reduce to a fourth. So we're going to get a final area under the curve of 1/8. If we look at another one of these, let's go from zero to one of the square root T to the fifth plus 2T times 5T to the 4th +2 DT. We're always going to think about trying to let our substitution be the most complex and or the composition what's inside of something else. So if I let U equal T to the fifth plus 2T, then DU is going to be 5T to the 4th +2 DT. We can see that this 5T to the 4th +2 DT is also equal to DU. So when we do a substitution here, we're going to get the square root of UDU. If I change my bounds, U is going to equal 0 to the fifth plus 2 * 0 for my lower bound, and U is going to equal 1 to the fifth plus 2 * 1 for my upper bound. So we're going to go zero to three. Now this is really just U to the 1/2. So if we add 1 to the exponent and divide by the new exponent, we'd get 2/3 U to the three halves from zero to three. So 2/3 EU to the three halves would be two to the three halves -0 to the three halves. Well, 2 to the three halves is really just two square roots of two, which is 4 root 2 / 3. Two to the three halves we could think of as two to the 1 + 1/2, because when we have an exponent, we can split it up into addition. And then if we have exponents that are added, that's the same thing as multiplying by the base so that the exponents get added. And then the 1/2 powers really sqrt 2, SO 4 sqrt 2 over 30. Except it wasn't a two, it was a three. Let's put in a three to the three halves. That changes our answer a little bit. Same idea, just not twos. It's threes instead. So 3 to the three halves we could think of as three to the 1 + 1/2, three to the 1 * 3 to the half, or three square roots of three. So three square roots of three. The threes then cancel and we get 2 square roots of three. If we look at some more examples, we're going to go into one that's a picture. If we have this one, we have a picture where we're given some information, IE 3 equations, and we're going to say let's find the area enclosed by this curve. So we're really trying to find this area in here, and we're going to do this in two different integrals. So we've got to figure out this area here to the left, and then we're going to figure out the area to the right. So when we look at the area to the left, we need to figure out what this X location is. Well, if we know that Y equal 1 for the horizontal line, and we also know Y equal X, we're really going to set those two equations equal to figure out what the X value is. And then this case X is 1. So this point right here would be the .11. So when we're looking at this integral, we're going to go from zero to one of the upper function. In this case, this is the top of those line segments, so the upper function is going to be X and we're going to subtract the lower function. In this case, the lower function is this curve X ^2 / 4. We're taking the integral in terms of the D axes because we're making very small slices in the X direction. Now that's only the area over here on the left. So now we have to figure out the area on the right portion. If we looked at over here next, so we know that our lower bound's going to be one. We had to figure out what this location is. So we're going to set those two equations equal and solve for X. So 4 equal X ^2 X is technically positive or -2, but because we're to the right of one, we know that we want 2. Then we're going to take the upper function, IE the top of the line, in this case 1 minus the lower function. The lower function in this case is X ^2 / 4. And this is all being done in terms of DX. So now we're going to evaluate this. So if we're trying to find the integral 0 to one of X, that's going to be a half X ^2 and minus X ^2 / 4, we're going to add 1 to the exponent and divide by the new exponent. So we get minus X ^3 / 12, and that's going to go from zero to 1 plus. Here the integral of one, it's a constant, so it's going to be X minus. Add 1 to the exponent X ^3. Divide by the new exponent 12 of one to two. Putting in the upper bounds we get 1/2 minus 112th. Putting in the lower bounds -0 -, 0 plus putting in the upper bound 2 -, 8 twelfths. Putting in the lower bounds 1 minus 112th. So simplifying this up, six twelfths minus 112th plus 24 twelfths -8 twelfths -1212 + 112. So let's see the -112 positive 1/12 cancel 6 + 20 four 3030 -, 2010 twelve 1012 is just 5-6. Now we're going to look at this exact same problem, but this time we're actually going to do it with an integration around the YS instead of the X's. So same exact picture, but this time we're going to look at doing it in terms of DY. If we thought about doing the DYS, we're going to be stacking horizontals. So what we need to figure out is we need to figure out our lower Y in this case, because it's the origin, the smallest our Y is a zero. Our upper Y, the highest the Y is ever going to go is one. Now we need to figure out this line segment length. So we're going to take the furthest to the right minus the furthest to the left. When we were doing verticals, we did the highest minus the lowest. Now we're doing the most positive or the furthest to the right minus the furthest to the left. So when we look at this, the furthest to the right is this Y equaling X ^2 / 4. But I don't want to integrate in terms of X this time. I want to integrate in terms of Y. So we're going to solve for X and we're going to get 4Y equal X ^2 or X equal positive or negative sqrt 4 Y. Now which one do we need? We're going to want the positive portion because we want XS to be positive. So two square roots of Y. So this equation, an equivalent just solving is X equal two square roots of Yi need my YS to be positive. I need my X's to be positive because we're located in that first quadrant. So we're going to have two square roots of Y minus. That's our furthest to the right. Now we need to figure out our furthest to the left in terms of Y. Well, this one was just Y equal X. So it's going to be plain old Y and we're integrating in terms of YDY. So now if we add 1 to the exponent, that square root of Y is Y to the 1/2. So we get three halves, and we're going to divide by the new exponent of three halves minus. Add 1 to the exponent, divide by the new exponent, and that's all going to go from zero to 1. So we're going to get 4 thirds Y to the three halves minus y ^2 / 2 from zero to 1. Substituting in the upper bound minus the lower bound 4 thirds minus 1/2, we're going to get 8 -, 3 / 6 or five sixths. Now this problem we could actually integrate in terms of X or Y. We're going to have some problems where we can't integrate in terms of both. If we look at this next example, we want to find the area that's enclosed by Y equal 1 and by Y equal cosine squared X. This is going to be an example where I have to do it in terms of DX's. I couldn't do it in terms of dy's because if I did, I'd go from the function back to the same function and that would cancel out. We can't do that. I have to have two unique functions. So the first thing we have to do is figure out where they intersect. So I'm going to set the two equal. If I square root, we're going to get positive and -1 cosine X. So when is cosine +1? Well, cosine is +1 at 0 and 2π. When is cosine -1 at π. So when we look at this curve, we're going to have at π cosine π is -1. Squared is 1, so here is going to be π and here was 0. So when we look at this integration, we're going to go from zero to π. Our upper bound is going to be one. Our lower bound is going to be cosine squared X DX. So when we look at this, we're going to have that equaling zero to Pi 1 minus cosine squared X is really the same thing as sine squared X DX. And currently we don't know how to integrate sine squared X DX. So we're going to use one of our trig identities that says sine squared X DX is a power reducing formula and it equals 1 minus cosine 2X over 2 DX. So if we integrate this, we get one half X -, 1/4 sine 2X from zero to π. So that derivative of 1/2 X was the 1/2 Here the derivative of negative 1/4 sine 2X the derivative of sine is cosine and the derivative of the two X would be two. So 1/4 * 2 would give us the 1/2. So now if we just stick in the upper bound minus the lower bound, we're going to see that we get 1/2 Pi minus sine of 2π, sine of 2π zero. 0 * 1/4 is 0 -, 0 sine of 0 is also zero. So we get a total of 1/2 Pi. And these are actually areas, so they all should be unit squared. For all of these problems that we've done, let's do one more. Let's do another one that's a bit more challenging. If we have Y equaling the absolute value of X underneath the square root, and the line 5 Y equal X + 6, we know that Y is going to be the square root of X when X is greater than or equal to 0, and Y is going to be the square root of the opposite of X when X is less than 0. So now we actually need to find where these intersect. So for the point over here on the right, we're going to have the square root of X equaling one fifth X + 6 fifths. Squaring each side. I'm going to get, oh, I'm going to actually multiply by 5 first to make life a little easier. So then when I square each side, I get 25 X equaling X ^2 + 12 X plus 36. Taking everything to one side, we get X ^2 -, 13 X plus 36. So zero is going to be X -, 9 * X -, 4 X is 9X is 4. So we're going to have it intersecting two different locations. It looks like this first one's going to be 4. And if we keep going out this way somewhere, we're going to have a nine. If we then look at on this equation over here, we're going to have square root of the opposite of X equaling one fifth X + 6 fifths. If I start by multiplying once again each side by 5. Now when I square it, we're going to get 25, but it's going to be -25 X because the square root and the square are going to cancel, but it's still going to be an opposite of X underneath. So X ^2 + 12, X plus 36. Taking everything to one side, we get X ^2 + 30, seven X + 36, or X plus One X + 36. So X equal -1 X equal -36 so here -1 and then if we kept going -36 I think that's going to actually be an extraneous root. I wonder if nine was an extraneous root. If we put 9 back into the original square root of, the absolute value of nine would be three. Over here. If we put nine in 9 + 6 is 15, so that one actually is a root. So both of these are roots. This one's got to be an extraneous root. So now what we're going to do is we're going to look at from -1 to 0 and then we're going to look at from zero to four. And then we're going to look at from 4:00 to 9:00 to get the total area. So when we look at from -1 to 0, we have to think about the upper minus the lower, so the upper in this case. And we're going to integrate in terms of X. So the upper is going to be one fifth X + 6 fifths minus the lower, and the lower in this case is going to be the negative no, the square root of the opposite of X. And that's all getting multiplied by DX. Then we're going to have that added to the integral, 0 to four. And on this integral, the top piece once again is going to be that line, one fifth X + 6 fifths minus the bottom, which is going to be the square root of X DX. Then from 4:00 to 9:00, the upper this time is going to be that square root of X. And then we're going to subtract the lower, which in this case is the line. OK, so if we looked out here from 4:00 to 9:00, we have the upper being the curve, the lower being the line. So now we're going to integrate all this. We're going to add 1 to the exponent and divide by the new exponent. If it's a constant, it's just the constant times X this next piece. This is negative X to the 1/2. So when we add one, we get three halves and we have to multiply or divide by the new exponent, which is 3 halves. So that'd be 2/3. But then we also have the derivative of the inside, which was a -1. So instead of a -, 2/3, it's going to turn to a + 2/3, and that's all going to go from -1 to 0. Plus we're going to add 1 to the exponent. Divide by the new exponent. This next part add 1 to 1/2 and get three halves divided by three halves, which is the same thing as multiplying by its reciprocal from zero to four. This next piece, add 1 to the exponent. Divide by the new exponent. So 2/3 X to the three halves minus 110th X ^2 -, 6 fifths X. Distributing the negative is what I did there in addition to finding the integration. So now if we put in the upper bounds, we can see the upper bound here of 0 is really just 0 minus. The lower bound -1 ^2 is positive, so we have a 110th -6 fifths. The opposite of -1 is 1, so plus 2/3 plus putting in the upper bound. Here 4 ^2 is 16, so 16 tenths plus 24 fifths minus sqrt 4 is 22 cubed is eight. 8 * 2 is 16 thirds minus the lower bound of 0, which is really just zero for all that plus the upper bound sqrt 9 is 3/3, cubed is 2727 * 2/3 would be 18. Then putting in the 9 here we get 9 ^2 81 tenths, putting in 9 * -6 fifths, so -54 fifths minus putting in four sqrt 4 is 22 cubed is eight, 8 * 2 is 16 thirds. Putting in 416 tenths, putting in 424 fifths. So if we combine all of our like terms, I believe we're going to get 5 thirds units squared. Thank you. And have a.