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[MUSIC].

Â Let's use calculus not just to calculate the area under the graph of a function,

Â let's use calculus to find the area between two curves.

Â Specifically, what's the area between the graph of y equals the square root of x.

Â And the graph of y equals x squared. As usual, let's start by drawing a graph

Â of the situation. Here's my plane.

Â the graph of y equals x squared maybe looks like that.

Â And the graph of the square root of x, maybe looks like this in red.

Â So what's this area between curves even mean?

Â Well I really mean the area inside, this little wing shaped region here.

Â this point is worth thinking about, that point has coordinates 1,1.

Â let's set it up as an integration problem.

Â I can imagine carving this region up into lots of thin rectangles, right?

Â Setting up a[UNKNOWN] sum. Let's draw some of those thin rectangles,

Â in this region here. And, you have to worry a little bit about

Â exactly how tall all these rectangles are.

Â And how wide all these rectangles are. Well, how tall are these rectangles?

Â Let's just pick one of the rectangles in particular to think about.

Â Let's pick this rectangle, and think about it.

Â How tall is it? Well, its top edge.

Â Is on y equals the square root of x, and its bottom edge is on this orange curve.

Â The y equals x squared term. So the height of that particular

Â rectangle is its top coordinate minus its bottom coordinate, so it will be square

Â root of x minus x squared tall And how wide is it?

Â Well, the width of this little tiny tri-, rectangle, is, not supposed to very big,

Â right? So let's call its width dx, right?

Â I'm imagining that its, width is, very narrow.

Â It's a very thin rectangle. So that amounts to evaluating a certain

Â integral. Yeah, but which integral?

Â Well, I want to add up the areas of all these little boxes.

Â And these little boxes have heights given by this, and widths given by some small

Â quantity, I'm[INAUDIBLE] dx. So the integral that I want to calculate

Â will be the integral of the area. So it'll be the height.

Â The square root of x minus x squared. Times the width which is DX.

Â And then what does my X go between? Well the smallest value of X here is zero

Â and the biggest value of X here is one so I'll have X going from zero to one.

Â So that's the integral that I want to calculate which will find the area in

Â between these two curves. I can evaluate that integral by using.

Â 2:59

The fundamental theorem of calculus. So, the fundamental theorem of calculus

Â tells me to find an anti-derivative for this.

Â So, let's first thing. What's an anti-derivative of the square

Â root of x? Well that's an anti-derivative of x to

Â the one half power. But remember my rule for

Â anti-differentiaing x to a power. It's x to that power plus one, so 3

Â halves is one more than a half, divided by that power plus 1, 3 halves so that

Â when I differentiate this I get back x to the one half power and I'll just add some

Â constant C on there to give myself the general anti-derivative.

Â What if I want to anti-differentiate x squared?

Â Well, that's maybe easier. And it's the same power rule.

Â It's x to the third power, which is one more than two, divided by two plus one.

Â So that when I differentiate this, I just get back x squared, and I'll add a

Â constant. Now what's an antiderivative of the

Â difference? Well, the difference of anti derivatives.

Â So, the aniti derivative of the square of x minus x squared is the anti derivative

Â for the squared of x which is x to the 3 halves over 3 halves, minus the anti

Â derivative for x squared. So, x to the 3rd over 3.

Â So, there's the general anti-derivative for this.

Â Armed with the anti-derivative, I can now evaluate the internal.

Â So, one way to write that is as follows. If I went to integrate from 0 to 1, the

Â squared of x, minus x squared dx. I'll just write down the anti-derivative

Â for this, or an anti-derivative for this. Which is x to the 3 halves over 3 halves,

Â minus x to the 3rd over 3. That's an anti-derivative for this, plus

Â 0, if you like. And then, I'll evaluate this quantity at

Â 0 and 1 and take the difference. So, this is a little bit of fancy

Â notation. Just for plugging in x equals 1, plugging

Â in x equals 0 and finding the difference. And if I want to emphasize what I'm

Â plugging in I might write even x equals 0 here just to emphasize that x is the

Â variable that I'm working with. Alright well let's do that.

Â let's plug in x equals 1 here. And when I do that I get 1 to the 3

Â halves power over 3 halves minus 1 to the third over 3.

Â And I'll subtract what I get when I plug in 0.

Â SO I plug in 0, 0 to the 3 halves over 3 halves minus 0 to the third.

Â Over three. Well this is easy to calculate now.

Â 1 to the 3 halves is 1. So it's the reciprocal of three halves.

Â Which is 2 3rds, minus 1 to the 3rd, which is 1 over 3.

Â So that's just a third. And here I've got minus 0.

Â So I've 2 3rds minus 1 3rd. This integral ends up being equal to 1

Â 3rd which is computing the area between the original curves.

Â I hope this problem wets your appetite for harder problems along these same

Â lines, right? You can imagine very complicated area

Â calculations.

Â