1 00:00:00,25 --> 00:00:05,35 [MUSIC]. 2 00:00:05,35 --> 00:00:09,655 Let's use calculus not just to calculate the area under the graph of a function, 3 00:00:09,655 --> 00:00:14,670 let's use calculus to find the area between two curves. 4 00:00:14,670 --> 00:00:20,120 Specifically, what's the area between the graph of y equals the square root of x. 5 00:00:20,120 --> 00:00:26,699 And the graph of y equals x squared. As usual, let's start by drawing a graph 6 00:00:26,699 --> 00:00:29,922 of the situation. Here's my plane. 7 00:00:29,922 --> 00:00:36,880 the graph of y equals x squared maybe looks like that. 8 00:00:36,880 --> 00:00:43,768 And the graph of the square root of x, maybe looks like this in red. 9 00:00:43,768 --> 00:00:47,300 So what's this area between curves even mean? 10 00:00:47,300 --> 00:00:53,732 Well I really mean the area inside, this little wing shaped region here. 11 00:00:53,732 --> 00:01:00,780 this point is worth thinking about, that point has coordinates 1,1. 12 00:01:00,780 --> 00:01:03,960 let's set it up as an integration problem. 13 00:01:03,960 --> 00:01:09,659 I can imagine carving this region up into lots of thin rectangles, right? 14 00:01:09,659 --> 00:01:14,84 Setting up a[UNKNOWN] sum. Let's draw some of those thin rectangles, 15 00:01:14,84 --> 00:01:18,271 in this region here. And, you have to worry a little bit about 16 00:01:18,271 --> 00:01:21,830 exactly how tall all these rectangles are. 17 00:01:21,830 --> 00:01:26,620 And how wide all these rectangles are. Well, how tall are these rectangles? 18 00:01:26,620 --> 00:01:29,470 Let's just pick one of the rectangles in particular to think about. 19 00:01:29,470 --> 00:01:31,770 Let's pick this rectangle, and think about it. 20 00:01:31,770 --> 00:01:34,650 How tall is it? Well, its top edge. 21 00:01:34,650 --> 00:01:40,470 Is on y equals the square root of x, and its bottom edge is on this orange curve. 22 00:01:40,470 --> 00:01:44,403 The y equals x squared term. So the height of that particular 23 00:01:44,403 --> 00:01:49,95 rectangle is its top coordinate minus its bottom coordinate, so it will be square 24 00:01:49,95 --> 00:01:53,972 root of x minus x squared tall And how wide is it? 25 00:01:53,972 --> 00:01:58,942 Well, the width of this little tiny tri-, rectangle, is, not supposed to very big, 26 00:01:58,942 --> 00:02:02,904 right? So let's call its width dx, right? 27 00:02:02,904 --> 00:02:06,831 I'm imagining that its, width is, very narrow. 28 00:02:06,831 --> 00:02:10,967 It's a very thin rectangle. So that amounts to evaluating a certain 29 00:02:10,967 --> 00:02:13,815 integral. Yeah, but which integral? 30 00:02:13,815 --> 00:02:16,790 Well, I want to add up the areas of all these little boxes. 31 00:02:16,790 --> 00:02:21,78 And these little boxes have heights given by this, and widths given by some small 32 00:02:21,78 --> 00:02:26,180 quantity, I'm[INAUDIBLE] dx. So the integral that I want to calculate 33 00:02:26,180 --> 00:02:31,218 will be the integral of the area. So it'll be the height. 34 00:02:31,218 --> 00:02:37,30 The square root of x minus x squared. Times the width which is DX. 35 00:02:37,30 --> 00:02:41,782 And then what does my X go between? Well the smallest value of X here is zero 36 00:02:41,782 --> 00:02:48,180 and the biggest value of X here is one so I'll have X going from zero to one. 37 00:02:48,180 --> 00:02:52,755 So that's the integral that I want to calculate which will find the area in 38 00:02:52,755 --> 00:02:58,643 between these two curves. I can evaluate that integral by using. 39 00:02:59,740 --> 00:03:03,220 The fundamental theorem of calculus. So, the fundamental theorem of calculus 40 00:03:03,220 --> 00:03:05,672 tells me to find an anti-derivative for this. 41 00:03:05,672 --> 00:03:08,636 So, let's first thing. What's an anti-derivative of the square 42 00:03:08,636 --> 00:03:11,690 root of x? Well that's an anti-derivative of x to 43 00:03:11,690 --> 00:03:14,696 the one half power. But remember my rule for 44 00:03:14,696 --> 00:03:19,115 anti-differentiaing x to a power. It's x to that power plus one, so 3 45 00:03:19,115 --> 00:03:23,455 halves is one more than a half, divided by that power plus 1, 3 halves so that 46 00:03:23,455 --> 00:03:28,75 when I differentiate this I get back x to the one half power and I'll just add some 47 00:03:28,75 --> 00:03:35,576 constant C on there to give myself the general anti-derivative. 48 00:03:35,576 --> 00:03:39,340 What if I want to anti-differentiate x squared? 49 00:03:39,340 --> 00:03:42,650 Well, that's maybe easier. And it's the same power rule. 50 00:03:42,650 --> 00:03:49,100 It's x to the third power, which is one more than two, divided by two plus one. 51 00:03:49,100 --> 00:03:52,516 So that when I differentiate this, I just get back x squared, and I'll add a 52 00:03:52,516 --> 00:03:55,330 constant. Now what's an antiderivative of the 53 00:03:55,330 --> 00:03:58,686 difference? Well, the difference of anti derivatives. 54 00:03:58,686 --> 00:04:04,783 So, the aniti derivative of the square of x minus x squared is the anti derivative 55 00:04:04,783 --> 00:04:10,243 for the squared of x which is x to the 3 halves over 3 halves, minus the anti 56 00:04:10,243 --> 00:04:18,312 derivative for x squared. So, x to the 3rd over 3. 57 00:04:18,312 --> 00:04:22,590 So, there's the general anti-derivative for this. 58 00:04:22,590 --> 00:04:26,622 Armed with the anti-derivative, I can now evaluate the internal. 59 00:04:26,622 --> 00:04:32,190 So, one way to write that is as follows. If I went to integrate from 0 to 1, the 60 00:04:32,190 --> 00:04:38,760 squared of x, minus x squared dx. I'll just write down the anti-derivative 61 00:04:38,760 --> 00:04:44,735 for this, or an anti-derivative for this. Which is x to the 3 halves over 3 halves, 62 00:04:44,735 --> 00:04:49,540 minus x to the 3rd over 3. That's an anti-derivative for this, plus 63 00:04:49,540 --> 00:04:53,692 0, if you like. And then, I'll evaluate this quantity at 64 00:04:53,692 --> 00:04:58,444 0 and 1 and take the difference. So, this is a little bit of fancy 65 00:04:58,444 --> 00:05:01,924 notation. Just for plugging in x equals 1, plugging 66 00:05:01,924 --> 00:05:06,668 in x equals 0 and finding the difference. And if I want to emphasize what I'm 67 00:05:06,668 --> 00:05:10,264 plugging in I might write even x equals 0 here just to emphasize that x is the 68 00:05:10,264 --> 00:05:14,624 variable that I'm working with. Alright well let's do that. 69 00:05:14,624 --> 00:05:20,740 let's plug in x equals 1 here. And when I do that I get 1 to the 3 70 00:05:20,740 --> 00:05:27,780 halves power over 3 halves minus 1 to the third over 3. 71 00:05:27,780 --> 00:05:30,540 And I'll subtract what I get when I plug in 0. 72 00:05:30,540 --> 00:05:37,138 SO I plug in 0, 0 to the 3 halves over 3 halves minus 0 to the third. 73 00:05:37,138 --> 00:05:40,742 Over three. Well this is easy to calculate now. 74 00:05:40,742 --> 00:05:45,270 1 to the 3 halves is 1. So it's the reciprocal of three halves. 75 00:05:45,270 --> 00:05:51,100 Which is 2 3rds, minus 1 to the 3rd, which is 1 over 3. 76 00:05:51,100 --> 00:05:56,30 So that's just a third. And here I've got minus 0. 77 00:05:56,30 --> 00:06:02,289 So I've 2 3rds minus 1 3rd. This integral ends up being equal to 1 78 00:06:02,289 --> 00:06:09,940 3rd which is computing the area between the original curves. 79 00:06:09,940 --> 00:06:13,468 I hope this problem wets your appetite for harder problems along these same 80 00:06:13,468 --> 00:06:22,308 lines, right? You can imagine very complicated area 81 00:06:22,308 --> 00:06:26,684 calculations.