1 00:00:00,281 --> 00:00:04,893 [MUSIC]. 2 00:00:04,893 --> 00:00:09,315 We dive into thinking more about volume and length and things, let's just focus a 3 00:00:09,315 --> 00:00:13,865 bit more on area. So let's find the area inside here. 4 00:00:13,865 --> 00:00:18,025 This orange curve is the graph y equals x squared, and this red line is just the 5 00:00:18,025 --> 00:00:23,710 horizontal line y equals 1. So I want to figure out the area of this 6 00:00:23,710 --> 00:00:29,135 region which is below the red line and above the orange curve. 7 00:00:29,135 --> 00:00:33,495 We already know how to attack a problem like this. 8 00:00:33,495 --> 00:00:41,130 I take this region and cut it up into a vertical strips. 9 00:00:41,130 --> 00:00:47,094 Alright, I'm going to imagine cutting it up into a whole bunch of thin rectangles. 10 00:00:47,094 --> 00:00:51,743 And then I just want to add up the areas of those rectangles integrate. 11 00:00:51,743 --> 00:00:56,000 Well to set up the intergral, and I should say pick a value of x. 12 00:00:56,000 --> 00:01:00,861 And think now, how tall is this thin rectangle? 13 00:01:00,861 --> 00:01:09,860 Well the top edge of the rectangle is at, 1, and the bottom edge is at x squared. 14 00:01:09,860 --> 00:01:13,210 So, the height of this rectangle, here say, is one minus x squared. 15 00:01:13,210 --> 00:01:18,807 That's the length here, say is 1 minus x squared. 16 00:01:18,807 --> 00:01:22,080 That's the length here. What's the width? 17 00:01:22,080 --> 00:01:24,620 Well, I'm just going to call that dx. Right, I'm going to imagine that these 18 00:01:24,620 --> 00:01:29,412 rectangles are real thin. So, now the area of just this one 19 00:01:29,412 --> 00:01:36,940 rectangle is 1 minus x squared, that's its height times its width dx. 20 00:01:36,940 --> 00:01:39,480 But I don't just want the area of this one rectangle. 21 00:01:39,480 --> 00:01:41,780 I want to add up the areas of all these rectangles. 22 00:01:41,780 --> 00:01:48,400 So, I'm going to integrate this from x goes minus 1 to 1. 23 00:01:48,400 --> 00:01:53,023 Alright, and I get those end points by thinkin about how, small and how large x 24 00:01:53,023 --> 00:01:58,035 can be in this region. So I put a minus 1 and a 1 there, and 25 00:01:58,035 --> 00:02:02,420 this definite interval will calculate the area of this region. 26 00:02:02,420 --> 00:02:04,180 And I can calculate that definitne integral. 27 00:02:04,180 --> 00:02:08,650 Let me just copy it down here. The integral that I want to calculate is 28 00:02:08,650 --> 00:02:13,580 the integral x goes from minus 1 to 1 of 1 minus x squared dx. 29 00:02:13,580 --> 00:02:16,310 And to calculate that integral, it's enough to use the fundamental theorem of 30 00:02:16,310 --> 00:02:20,192 calculus. So I'll write down an anti-derivitive. 31 00:02:20,192 --> 00:02:25,232 x minus x cubed over 3 is an anti-derivitive and then the fundemental 32 00:02:25,232 --> 00:02:30,104 theorum of calculus says evaluate this at 1 and at minus 1 and I take the 33 00:02:30,104 --> 00:02:36,134 difference. So, I will plug in 1 and I'll get one 34 00:02:36,134 --> 00:02:41,326 minus 1 cubed over 3 and I am going to subtract what I get when I plug in minus 35 00:02:41,326 --> 00:02:47,713 1 which minus 1 minus, minus 1 cubed over 3. 36 00:02:47,713 --> 00:02:53,355 But what's this, this is 1 minus a third that's 2 3rd minus, minus 1 plus a 3rd 37 00:02:53,355 --> 00:03:01,160 which is negative 2 3rd, and 2 3rd minus negative 2 3rd, is 4 3rd. 38 00:03:01,160 --> 00:03:05,960 So the area of this region is 4 3rd square units, but I could also do this by 39 00:03:05,960 --> 00:03:11,080 cutting the region up into horizontal strips. 40 00:03:11,080 --> 00:03:18,148 So, here I am cutting this region up into some horizontal rectangles, which instead 41 00:03:18,148 --> 00:03:23,635 of being thin in terms of their width, they're now thin in terms of their 42 00:03:23,635 --> 00:03:29,728 height. I just want to add up all of these not 43 00:03:29,728 --> 00:03:35,290 very tall rectangles to compute the area of this region. 44 00:03:35,290 --> 00:03:39,378 So to do that I'm going to pick some value of y and I'm going to think about, 45 00:03:39,378 --> 00:03:44,685 for that value of y, how wide is is that rectangle? 46 00:03:44,685 --> 00:03:47,870 Well let's think about this point over here, right? 47 00:03:47,870 --> 00:03:52,198 This is the curve y equals x squared. So for this value of y what's the 48 00:03:52,198 --> 00:03:57,376 corresponding value of x? Well it's the square root of y, and that 49 00:03:57,376 --> 00:04:02,460 tells me how wide this whole rectangle is right. 50 00:04:02,460 --> 00:04:06,140 And this point over here is negative the square root of y. 51 00:04:06,140 --> 00:04:10,604 So, from here to here, is 2 square roots of y, that's the width of, of this 52 00:04:10,604 --> 00:04:16,482 rectangle, let me write that down. So, I got the width of the rectangle is 2 53 00:04:16,482 --> 00:04:20,985 square roots of y. Now, how tall is that rectangle? 54 00:04:20,985 --> 00:04:24,633 Well, let's call that dy, right, I'm imagining that the rectangle isn't very 55 00:04:24,633 --> 00:04:29,926 tall, really thin. So the product 2 squared of y its width, 56 00:04:29,926 --> 00:04:35,909 and it's not very tall height d y. That gives me the area of one of these 57 00:04:35,909 --> 00:04:39,950 thin rectangles and I'm going to integrate that. 58 00:04:39,950 --> 00:04:43,950 But from where to where? Well, y could be as small as 0 and as big 59 00:04:43,950 --> 00:04:49,274 as 1. So this integral will calculate the area 60 00:04:49,274 --> 00:04:54,480 of this region, decomposed into horizontal strips. 61 00:04:54,480 --> 00:04:57,020 I was going to do that integral, but I can definitely do that, right? 62 00:04:57,020 --> 00:04:59,790 I can do this integral by using the fundamental theory of calculus. 63 00:04:59,790 --> 00:05:04,350 I just copy that integral down over here. It would be integrating from 0 to 1, 2 64 00:05:04,350 --> 00:05:10,776 square roots of y dy, l only rewrite that as the interval from 0 to 1 of 2y to the 65 00:05:10,776 --> 00:05:16,293 1 half power dy. And now I can write down an 66 00:05:16,293 --> 00:05:19,080 anti-derrivitave of this by using the power rule. 67 00:05:19,080 --> 00:05:25,262 So an anti-derrivitave is 2y to the 3 halves power, divided by 3 halves. 68 00:05:25,262 --> 00:05:29,090 Right, an anti-derivative y to a power, is y to one more than that power, divided 69 00:05:29,090 --> 00:05:33,580 by one more than that power. I want to evaluate that at 0, and at 1, 70 00:05:33,580 --> 00:05:39,658 and take the difference. Well, when I plug in 1, I get 2 over 3 71 00:05:39,658 --> 00:05:44,166 over 2. When I plug in 0 I get 0, so the answer's 72 00:05:44,166 --> 00:05:48,982 2 divided by 3 over 2, which I can rewrite as 4 3rd, which is the area of 73 00:05:48,982 --> 00:05:54,978 this region. The neat thing here is that either way, 74 00:05:54,978 --> 00:05:58,695 we're getting the same answer, and I think this is more amazing than it seems 75 00:05:58,695 --> 00:06:02,542 at first. I mean look, when I cut it into 76 00:06:02,542 --> 00:06:06,763 horizontal strips I ended up wanting to calculate this interval, the interval y 77 00:06:06,763 --> 00:06:12,590 goes from 0 to 1 2 square roots of y dy. And when I cut the region up into these 78 00:06:12,590 --> 00:06:16,181 vertical strips I wanted to do the interval x goes from minus 1 to 1 of 1 79 00:06:16,181 --> 00:06:21,580 minus x squared dx. But these two intervals end up being 80 00:06:21,580 --> 00:06:26,136 equal because they're both calculating the area of the same region, they're both 81 00:06:26,136 --> 00:06:29,870 calculating 4 over 3 square units.