1
00:00:05,310 --> 00:00:06,649
[MUSIC]. 
Let's find the area between two 

2
00:00:06,649 --> 00:00:11,161
parabolas. 
I want to find the area between the graph 

3
00:00:11,161 --> 00:00:18,980
y equals x squared and the graph of y 
equals 1 minus x squared. 

4
00:00:18,980 --> 00:00:23,260
As usual let's start by graphing. 
Well here's my coordinate plane. 

5
00:00:23,260 --> 00:00:32,460
I'll draw a graph of y equals x squared. 
And the other curve is y equals 1 minus x 

6
00:00:32,460 --> 00:00:39,170
squared, so I'll draw a downward-facing 
parabola. 

7
00:00:39,170 --> 00:00:42,980
there's the one side of it and there's 
the other side of it. 

8
00:00:42,980 --> 00:00:47,460
I'll draw some rectangles to get an idea 
about what Riemann Sum I really want to 

9
00:00:47,460 --> 00:00:52,58
take the limit of. 
So it's really the area inside here that 

10
00:00:52,58 --> 00:00:56,972
I want to calculate. 
Draw some some rectangles then for the 

11
00:00:56,972 --> 00:01:04,116
for the Riemann Sum. 
That would approximate that area. 

12
00:01:04,116 --> 00:01:08,518
Right, so there's some rectangles for the 
the Riemann sum that would approximate 

13
00:01:08,518 --> 00:01:12,740
the area in there. 
And then I can write down the heights and 

14
00:01:12,740 --> 00:01:17,666
the widths of one of those rectangles. 
Let's take a look and say this rectangle 

15
00:01:17,666 --> 00:01:20,456
here. 
Well, what's the width of that rectangle, 

16
00:01:20,456 --> 00:01:23,660
will be real thin, so I'll call the width 
dx. 

17
00:01:23,660 --> 00:01:27,488
And what's the height of that rectangle. 
Well the top of that rectangle is on the 

18
00:01:27,488 --> 00:01:30,908
curve 1 minus x squared and the bottom of 
that rectangle was on the curve x 

19
00:01:30,908 --> 00:01:34,349
squared. 
So if we take the top coordinate minus 

20
00:01:34,349 --> 00:01:38,355
the bottom coordinate, that will give me 
the height of this rectangle. 

21
00:01:38,355 --> 00:01:43,424
The top coordinate is 1 minus x squared. 
The bottom y coordinate is just x 

22
00:01:43,424 --> 00:01:46,974
squared. 
So this quantity is giving you the height 

23
00:01:46,974 --> 00:01:51,460
of this rectangle. 
Oh, now I can write down the integral. 

24
00:01:51,460 --> 00:01:55,0
So I know the height and the width of 
these rectangles so the thing I want to 

25
00:01:55,0 --> 00:01:58,840
add up are the areas of those little 
rectangles. 

26
00:01:58,840 --> 00:02:04,166
So I want to integrate the height. 
1 minus, this is 2x squared, times the 

27
00:02:04,166 --> 00:02:08,149
width, dx. 
But I'm trying to integrate from where to 

28
00:02:08,149 --> 00:02:12,136
where, right? 
What are the possible x values in this 

29
00:02:12,136 --> 00:02:15,630
region? 
Well, if we think about exactly where 

30
00:02:15,630 --> 00:02:20,677
these two curves cross, right? 
What's the x coordinate where these 

31
00:02:20,677 --> 00:02:24,689
curves cross? 
And this coordinate here is minus 1 over 

32
00:02:24,689 --> 00:02:29,830
the square root of 2 and this coordinate 
is 1 over the square root of 2. 

33
00:02:29,830 --> 00:02:34,318
So I'm going to be integrating to compute 
the area from x equals negative 1 over 

34
00:02:34,318 --> 00:02:39,148
the square root of 2 up to 1 over the 
square root of 2. 

35
00:02:39,148 --> 00:02:44,710
And it's this integral that'll calculate 
the area in between those two curves. 

36
00:02:44,710 --> 00:02:49,630
I can use the fundamental theorem of 
calculus to evaluate this integral. 

37
00:02:49,630 --> 00:02:53,469
Let's find an antiderivative. 
So what's an antiderivative of 1, well x 

38
00:02:53,469 --> 00:02:59,6
is an antiderivative of 1. 
What's an antiderivative of 2 x squared, 

39
00:02:59,6 --> 00:03:05,440
well 2 x to the 3rd over 3 differentiates 
to 2 x squared. 

40
00:03:05,440 --> 00:03:10,404
And I'm looking for an antiderivative of 
a difference, so it'll be the difference 

41
00:03:10,404 --> 00:03:15,319
of antiderivatives. 
Now I'll evaluate my antiderivative at 

42
00:03:15,319 --> 00:03:19,490
the right and left endpoints. 
Right, so I have to evaluate my 

43
00:03:19,490 --> 00:03:23,650
antiderivative at the end points, which 
are 1 over the square root 2 and minus 1 

44
00:03:23,650 --> 00:03:28,694
over the square root of 2 and then take 
the difference. 

45
00:03:28,694 --> 00:03:32,552
So what do I get when I plug in 1 over 
the square root of 2. 

46
00:03:32,552 --> 00:03:37,450
I get 1 over the square root of 2 minus 2 
times 1 over the square root of 2 to the 

47
00:03:37,450 --> 00:03:42,966
third divided by 3. 
So this is what I get when I plug in x 

48
00:03:42,966 --> 00:03:48,830
equals 1 over the square root of 2 into 
my antiderivative. 

49
00:03:48,830 --> 00:03:53,822
And I subtract what I get when I plug in 
negative that, which is negative 1 over 

50
00:03:53,822 --> 00:03:58,268
the square root of 2, minus 2 times 
negative 1 over the square root of 2 

51
00:03:58,268 --> 00:04:04,286
cubed, all over 3. 
I can simplify this a bit. 

52
00:04:04,286 --> 00:04:09,580
Well, first off, I've got a quantity 
minus negative that same quantity. 

53
00:04:09,580 --> 00:04:13,96
That's just two copies of this same 
quantity. 

54
00:04:13,96 --> 00:04:15,676
Now I can keep calculating this and try 
to simplify this even a little bit 

55
00:04:15,676 --> 00:04:17,956
further. 
Now let's keep going. 

56
00:04:17,956 --> 00:04:21,990
So what do I have here? 
Well it's 2 times 1 over the square root 

57
00:04:21,990 --> 00:04:25,762
of 2, minus 2. 
And here I've got 1 over the square root 

58
00:04:25,762 --> 00:04:29,971
of 2 to the 3rd power. 
So I could write that as a half. 

59
00:04:29,971 --> 00:04:35,174
Times 1 over the square root of 2. 
That's really 3 copies of the square root 

60
00:04:35,174 --> 00:04:37,905
of 2 divided by 3. 
Okay. 

61
00:04:37,905 --> 00:04:44,460
But, 2 times 1/2, alright, cancels. 
And then I've got a common factor of 1 

62
00:04:44,460 --> 00:04:48,875
over the square root of 2. 
So I could, pull out that common factor 

63
00:04:48,875 --> 00:04:54,30
of 1 over the square root of 2. 
And when I pull that out I've got a one 

64
00:04:54,30 --> 00:04:59,723
there minus a third is left. 
now 2 times 1 over the square root of 2, 

65
00:04:59,723 --> 00:05:04,826
that's just the square root of 2 and 1 
minus a third well that's really 2 3rds, 

66
00:05:04,826 --> 00:05:11,796
so what's left all over here is the 
square root of 2. 

67
00:05:11,796 --> 00:05:17,428
Times 2 3rds, this is the value of the 
integral and also then, the area between 

68
00:05:17,428 --> 00:05:22,233
those two curves. 
Before committing to this as my final 

69
00:05:22,233 --> 00:05:26,701
answer, I should check to see that the 
answer's somewhat reasonable. 

70
00:05:26,701 --> 00:05:32,926
Well if we numerically approximate this, 
this is about 0.94. 

71
00:05:32,926 --> 00:05:38,200
So is a bit less than one square unit a 
reasonable answer? 

72
00:05:38,200 --> 00:05:43,895
Well the region that we're interested in 
fits inside a rectangle of height 1, and 

73
00:05:43,895 --> 00:05:52,527
width the square root of 2 units. 
The area of this rectangle is about 1.41 

74
00:05:52,527 --> 00:05:57,523
square units. 
And the area of the region we just 

75
00:05:57,523 --> 00:06:02,20
calculated to be about 0.94 square units, 
right? 

76
00:06:02,20 --> 00:06:04,730
This region inside here. 
And it's pretty reasonable. 

77
00:06:04,730 --> 00:06:07,460
I mean, you know? 
It looks like. 

78
00:06:07,460 --> 00:06:14,93
If this thing is about 1.41 square units 
it's not unreasonable to think that this 

79
00:06:14,93 --> 00:06:22,416
is a little bit less than 1 square unit 
in this curved region, here. 

80
00:06:22,416 --> 00:06:29,993
Yea, we're really finding answers that 
accord with our geometric intuition.