 Welcome back everyone. Let's learn some more about calculus two specifically. Let's do some more examples to find an area between the curves In this example, we want to find the area between the curve y equals x to the one-half power That is the usual square root function You can see that in front of you illustrated in yellow right here And let's also find the area between the square root function and the then the function y equals x cubed Your standard cubic function, which you can see illustrated here in green Let's call Let's call because sometimes it's nice to give them names here instead of y equals to square root We'll call this f of x equals to square root and instead of y equals x cube. We'll call that g of x Equals x cube right there and so you can see this illustrated in the picture now unlike some of the previous example. We saw here They didn't tell you what the bounds are here. What's the boundary of this problem x equals a x equals b What are my a and b value? They didn't tell you these bounds because it's implicit in the description of these functions We're looking for a finite region who has a finite area And when you look at the graph of these two functions the square root of x and x cubed We can see that there's only one Natural region bounded by these two functions which you can see illustrated here in light blue And so we're trying to find the area of this region And so even though they didn't tell us what the bounds are we can infer from the description of the region What the bounds are going to be and so again this can be helpful when we have the picture in front of us You could try drawing yourself or using a graphing calculator. If you don't have a graphing calculator You can always use desmos.com It's a free graphing Calculator works pretty well actually that you might we'll actually probably see it in some future videos as well Desmos.com it offers a free graphing calculator. Feel free to use it in your studies in your homework I think it'll be a great use to you here So we can see from this picture that it seems like there's an intersection right here I would guess from the graph that's going to be zero zero the origin And it also appeared that there's some other intersection right here by my graph It looks like one one and you could check that these these two points are in fact points of intersection But what if you didn't know where they intersected what if your drawing wasn't very accurate or you struggled to find them You could always find these intersections algebraically and this is the preferred method I would suggest to find here that if you want to find the intersection Assuming you can spell the word if you want to find the intersection of two functions The idea is to set them equal to each other f of x equals g of x set them equal to each other and solve for x Well f of x remember was the square root of x g of x was x cubed In order to make this an easier equation to solve I prefer to have polynomial a polynomial function So we're gonna square both sides square the left hand side to get rid of the square root and square the right hand side Because what's good for the goose is good for the gander a Gander's just a boy goose in case you didn't know what that meant the square root of x squared is an x x cubed squared You're actually gonna multiply the exponents together and so we get x the sixth And so setting this polynomial equal to zero we get x to the sixth minus x equals zero Proceed I mean you could try to factor this thing if you want to x you can factor out an x Then you're gonna get x minus one And then likes an x to the fourth plus x cubed plus x squared plus x plus one is Equal to zero if I spazzed everyone out right there. That's my apologies, right factoring can be a little bit more challenging here Some of us might be more inclined to try this by division if we divide both sides by x We end up with one equals x to the fifth now be aware that if you divide by x you're assuming x isn't equal to zero But x actually is a solution a zero is a solution right here So you divide by x make sure you keep track that x equals zero as a possibility in fact that is one of the points of intersection We saw on the graph over here, right x equals zero and then if you have one equals x to the fifth we'll take the fifth root of both sides Do that to the other side as well and you're gonna see that x equals one is the other real solution And that's gonna correspond to this one right here So we found our bounds x equals zero and x equals one It takes a little bit more effort to find them because it was implicit with the description But once we have it we're ready to go to find the area between the curves We're gonna integrate from zero to one and we're gonna take the bigger function minus the smaller function So we take x to the one half We see this from the picture that the square root is the bigger function and subtract from it x cubed like so DX now before continuing I do want to make a comment here What if what if I botched it? What if I put x cubed in front of x to the one half and I because I thought that was the bigger one What's gonna happen? Well by reversing by reversing the roles of x cubed and x to the one half We actually see here that We would just get the same area the same region, but the area is negative So if you get a negative answer at the end of these problems just take the absolute value Don't fix a too much on making sure you get the right order But we do we do prefer to put the bigger one first, right? So taking the anti-derivative we use the power rule right here the one half power will increase by one it becomes the three halves power and then divide by three halves the new exponent and Then x cubed it'll upgrade to be x to the fourth over four and then let's go from zero to one right here Now if you divide by the fraction three halves, that's the same thing that's multiplied by its reciprocal So you all more often prefer to write this as two-thirds multiplication here x to the three halves Minus x to the fourth over four right from zero to one You'll see that these bounds are actually quite benign right when you plug in zero because you have only powers of x everything will disappear That's really nice and when you plug in the one because powers of one are always equal to one This thing will simplify to be two-thirds minus a fourth which as far as arithmetic's go is is pretty benign, right? We do do write the common denominator here between three and four and your least common multiple we 12 So we'll times the first one four over four and then the second is to be three over three So we end up with eight minus three twelfths and So then we find out the area between the term the between the the two curves is five twelfths Let's take a look at another example here Let's find the area enclosed by the line y equals x minus one and the parabola y squared equals 2x plus 6 So again much like the last example They don't tell you the bound x equal a equal b because it's implicit by the description here If we were to graph these things you see in green the function y equals x minus one the line I should say y equals x minus one right here And then in yellow you see the parabola y squared equals 2x plus 6 You'll notice that the the yellow curve is it not a function in the standard meaning right? It fails the vertical line test and that comes to the fact that you have a y squared as opposed to an x squared This thing is a concave right Parabola and that's actually not too much of a problem for us We could try I mean if we wanted to we could solve for y just by taking taking the square root But the issue with that is if we take the square root We actually get the positive square root and the negative square root So we'd have to treat this as like a piecewise function and break it up into two integrals Which might not necessarily be the most preferable technique On the other hand when we first talked about this area between the curve We have this f of x minus g of x dx But there's nothing that actually forces us to use the differential dx right dx gives us these Vertical rectangles like this on the other hand we could switch our focus and focus instead on Horizontal rectangles like so if we do these horizontal rectangles Then the thickness of this rectangle will actually be the differential dy So I guess what I'm saying is you are to find areas under curves between curves and things like that You were welcome to integrate with respect to y instead of with respect to x if it's a little bit easier The reason that would be a better choice in this situation is that for the for the parabola if you solve for x You'll just subtract this 6 and divide both sides by 2 we get that x equals y squared minus 6 over 3 Not over 3. I'm sorry over 2 Which I would simplify to be one half y squared minus 3 Which is a pretty good quadratic function for x and also this guy right here You can solve for x in which case you get x equals y plus one That's fairly painless right there So we can describe both of these functions in terms as x as a variable y instead of the usual y as a variable of x And so if we do that way we have to we have to kind of change our perspective over here Which function is bigger than the other one and that's going to be the one on the right Because the farther along the x-axis you get the more positive things are so in this situation The green function is the bigger function than the yellow one So we're going to take this to be we'll call the green one f of x We'll call the yellow one g of x and so we're going to take f minus g Define the area between the curves, but we have to also figure out where the boundaries points right here We can try to see them from the graph But again, let's try to see it algebraically what these things are You would set the two functions equal to each other y squared minus six over two Is equal to y plus one? I think I'm going to distribute the fraction again, so we'll take one half y squared minus three like so Let's times both sides by two. I guess no matter what I do. I don't want the fraction there time It's both sides by two so we get y squared minus six is equal to two y plus two If you are at all afraid of fractions that as you suffer from a debilitating disease called ratio phobia the fear fractions This is a support group for you. We love you We care about you and nearly always we can get rid of fractions in our arithmetic just by multiplying both sides of the equation by The denominator we want to get rid of this is now a quadratic equation move everything to the same side I'm going to subtract two y and subtract two from both sides This gives us y squared minus two y minus eight equals zero We then want to factor this thing or we could use the quadratic formula Whichever you feel more comfortable with the complete the square is also an option But we want factors of negative eight that add up to be negative two and with a very quick inspection We can see that y equals negative four and y plus two does the trick Right negative four times two gives us negative eight negative four plus two gives us negative two And so we see the two points of intersection are y equals positive four y equals negative two Which seems to look like the values we see right here I mean if I were to guess from the graph one two three four five the x-cord looks like it's five The y-coordinate I did find the y-coordinate didn't I? Four and negative two. Yeah, that's not that's not a problem there Where is my issue here? This one right here? I wonder if this nice my scales probably off on my y-axis Whoops a daisy. That's probably the issue. Don't look at that Embarrassing but see this this is again what emphasizes the point why I say focus on the algebra The geometry is good to inform our intuition But we should rely more on the algebra because we're less inclined to make mistakes such as the one you see in the video here Anyways, so now that we know our bounds. Let's get ready to find the area the area between the curves We're gonna take the integral which remember before we said we're gonna take F minus g and so the bigger function was the linear function y plus one and then we're gonna subtract from it The quadratic function one half y squared Minus three was it a minus three or plus three. I've already forgotten. It was a minus three Like so we're gonna integrate with respect to y And so next thing to pay attention to are the boundaries, right? This is what we had to find this y-coordinates if you're integrating with respect to y the bounds over here need be y Coordinates not x coordinates y will range from negative two to four even though x will range from like Negative one to five or something like that We want to make sure we get the y coordinates the boundary of the integral needs to match the differential Just like the variable we integrate here needs to match this thing right here Many students sometimes ignore the differential thinking it's very insignificant, but that's not true at all We actually need that differential to help us First of all, it's a geometric ops Geometric measurement is keeping track of the thickness of our rectangles But also it helps us make sure we have everything correct inside of our Inside of our function right here. So let's then proceed to integrate Let's combine some like terms and distribute that negative sign inside of the integral So we're gonna get integral from negative two to four We're gonna get negative one half y squared Plus y and so the only thing that combines is the constant we get one plus three notice how it's a double negative We distribute the negative sign there. So we get a four and so integrate this by the usual power rule We would end up with Negative one sixth y cubed so I divided by three and so since I already had a one half divided by three It's just as it's the same thing. It's just times you by one third So you get one half times a third, which is a sixth negative one six y cubed plus y squared over two Plus four x for y. Excuse me That is a four There we go, and then we go from negative two to four and so at this moment the arithmetic becomes extremely hideous By all means use the scientific calculator a calculator of some kind to help you out right here No one has to be a hero. No one's gonna applaud you for your amazing arithmetic abilities here, right? It's number crunching at this moment plug in the four So you get four cubed which is 64 plus four squared over two. I forgot to put in the four Four squared over two it's 16 over two. That's gonna be an eight plus four times four. That's another 16 Subtract from that when we plug in the negative two. We're gonna get positive eight over six We're gonna get positive four over two and then minus eight Again, there's a lot going on there. This will simplify. I'm not gonna bother us with the arithmetic here You can simplify this a little bit more And which case this then simplifies to be something like 18 you can double-check the arithmetic yourself The calculus part of course is the part we really care about for a class like this And so we get the area under area between this curve is gonna be 18 All right, and so we saw on this video how even if they don't tell us the boundary between Two curves we can still identify it by setting the two functions equal to other and solving it All right, I'll see you guys next time