00:0-1 | No . In this video , we're going to talk | |

00:02 | about how to calculate the area between two curves . | |

00:06 | So let's go over the basics , let's say we | |

00:09 | have some function F of X . And we want | |

00:14 | to find the area under the curse from A to | |

00:20 | B . So we're looking for the area of the | |

00:28 | shaded region . The area is simply the definite integral | |

00:33 | from A to B of F . Of X . | |

00:37 | Dx . Now , let's say if we have another | |

00:42 | function from A to B and let's call this function | |

00:50 | G of X . The area under this curve is | |

01:00 | going to be the integral from A to B . | |

01:04 | But of G of X . Dx . Now , | |

01:08 | what happens if we want to find the area between | |

01:10 | two curves ? So let's say if we have both | |

01:15 | F of X , let me draw this better N | |

01:21 | . G fx . How can we find the area | |

01:30 | between these two curves ? All we need to do | |

01:38 | is take the difference between this area in this area | |

01:43 | and we'll get the area between F and G of | |

01:46 | X . So that area is simply the integral of | |

01:51 | a to b half of axe minus G fx dx | |

01:59 | . So that's how you can find the area between | |

02:01 | the two curves . You need to take the difference | |

02:04 | between the top function and subtracted by the bottom function | |

02:07 | and then take the definite integral of that difference and | |

02:11 | you'll get the area of that curve . Now let's | |

02:16 | see if we have another function . We'll call this | |

02:20 | function F of Y . Let's say it varies from | |

02:25 | C two d along the y axis . And we | |

02:32 | want to calculate the area from C . Diddy between | |

02:38 | that curve and the Y axis . The area of | |

02:42 | that region is going to be the integral from C | |

02:47 | Diddy of F of Y Dy . So that function | |

02:55 | is basically X is equal to some function of Y | |

03:03 | . Now , let's see if we want to find | |

03:06 | The area between two curves . So let's say we | |

03:11 | have F . Of Y and G F Y going | |

03:25 | from C . Diddy . So instead of taking the | |

03:33 | top function and subtracting by the bottom function , what | |

03:36 | we're gonna do is we're gonna take the integral from | |

03:39 | C to D . Of the function on the right | |

03:43 | , which is F . Of y minus the function | |

03:49 | on the left , which is G . F . | |

03:54 | Y . And so that's how we can get the | |

03:57 | area between two curves from C . Diddy . When | |

04:01 | using by values , I remember when dealing with F | |

04:06 | of X , it's Y is equal to fx . | |

04:09 | When dealing with F F Y , it's X is | |

04:12 | equal to F F Y . Now let's go ahead | |

04:17 | and work on some practice problems calculate the area of | |

04:20 | the region bounded by the line . Y equals 8 | |

04:24 | -2 x . The X axis and the Y axis | |

04:30 | . Well , let's begin by drawing a picture . | |

04:33 | So y equals 8 -2 . X . Let's get | |

04:38 | the X and Y intercepts . Since we're dealing with | |

04:40 | a linear equation when X is zero , this is | |

04:44 | going to be eight minus two times zero , which | |

04:45 | is eight . And if we replace why with zero | |

04:50 | and solve for X , we can see that X | |

04:54 | Is going to be eight divided by two , which | |

04:56 | is four . So we have a y intercept of | |

05:02 | eight And the x intercept of four . So we | |

05:17 | want to find the area between this line , the | |

05:21 | X axis and the Y axis . So one way | |

05:30 | we can do this is through the use of geometry | |

05:33 | . We know that the area of a triangle is | |

05:35 | one half base times height . The basis for the | |

05:40 | highest eight half of two is , I mean half | |

05:43 | of force to Times eight , that's 16 . So | |

05:47 | that's the area of the shaded region . But let's | |

05:50 | use calculus to get this answer as well . So | |

05:54 | another area between a curve and the X axis is | |

05:58 | going to be the integral from A to B . | |

06:00 | Of F . Of X . Dx . And as | |

06:04 | you know , why is equal to F . Of | |

06:06 | X And why is 8 -2 x . So FFX | |

06:12 | can be replaced with 8 -2 x . And along | |

06:16 | the X axis . We're going to integrate it from | |

06:18 | 0 to 4 . So we're gonna have the integral | |

06:25 | Of or from 0 to 4 . Activex , which | |

06:28 | is 8 -2 x . And then dx to the | |

06:34 | anti derivative of eight is eight X . And the | |

06:37 | anti derivative of two X . Of the first power | |

06:40 | . It's two X squared over two so we can | |

06:44 | cross altitude . So it's just eight x minus X | |

06:47 | squared . So now let's plug in four . So | |

06:51 | it's going to be eight times four minus four squared | |

06:55 | and then we'll plug in zero . So it's eight | |

06:57 | times zero minus zero squared . eight times 4 is | |

07:02 | 32 . 4 squared is 16 . And everything on | |

07:06 | the right will be zero And 30 to -16 is | |

07:09 | 16 . So we get the same answer and uh | |

07:15 | it's confirmed . So that's how you can calculate the | |

07:17 | air of the region using this particular definite integral . | |

07:21 | Everything Now we can also get the same answer in | |

07:27 | terms of why as opposed to in terms of acts | |

07:32 | so we can integrate it from C to D . | |

07:37 | Using this formula . Remember X is equal to F | |

07:41 | . Of Y . So what we need to do | |

07:45 | to get a function of why is we need to | |

07:47 | solve for X . So we have Y . is | |

07:50 | equal to 8 -2 . X . I'm going to | |

07:54 | move this to the left side where it's gonna be | |

07:56 | positive two X . I'm gonna move wide to the | |

07:59 | right side . War it's going to be negative Y | |

08:03 | . And now let's divide everything by two . So | |

08:07 | we get X is equal to four -1/2 y . | |

08:16 | So therefore we can say that f of Y is | |

08:20 | equal to four minus one halfway . We're going to | |

08:28 | integrate it from C . to D . or 0 | |

08:31 | - eight . So the area is going to be | |

08:33 | the integral from 0-8 . This is why which is | |

08:37 | for -1/2 . Why ? And then dy So the | |

08:44 | anti derivative of four , it's going to be for | |

08:45 | Y . The anti derivative of why to the first | |

08:48 | power is why it's the second power over to Evaluated | |

08:53 | from 0 - eight . So this is four y | |

08:57 | minus 1/4 times Y squared . So let's plug in | |

09:03 | eight . We're gonna have four times 8 -1 4th | |

09:07 | times eight squared . And then once we plug in | |

09:10 | zero into this expression , the whole thing is going | |

09:13 | to be zero . So four times 8 is 32 | |

09:20 | And then eight square is 64 64 times 1/4 is | |

09:25 | 16 . So this too will give us the same | |

09:28 | answer of 16 square units for the area . So | |

09:33 | regardless if we choose to find the area , in | |

09:36 | terms of X , or in terms of why we're | |

09:38 | gonna get the same answer , let's work on this | |

09:41 | problem , calculate the area of the region bounded by | |

09:44 | the line Y equals X and Y equals x squared | |

09:52 | y equals X . Is basically a straight line that | |

09:54 | passes through the origin At an angle of 45°. . | |

10:02 | That's why equals x , Y equals ax squared , | |

10:08 | is basically a problem that opens upward but starts at | |

10:13 | the origin . So if we put those two graphs | |

10:16 | together , we're gonna get something that looks like this | |

10:28 | . So here's the line Y equals X . And | |

10:31 | here is why equals X squared on the right side | |

10:36 | , on the left side . It just looks like | |

10:37 | that , but we don't need to worry about it | |

10:40 | . The area of the region bounded by these two | |

10:43 | curves is right here . So that's the area that | |

10:47 | we need to calculate . What we need to do | |

10:51 | is we need to determine the points of intersection . | |

10:57 | We can see the first point is going to be | |

10:58 | at zero . We got to find the second point | |

11:01 | to do that . We want to set these two | |

11:03 | equal to each other , so we want to set | |

11:06 | X equal to X squared . Subtracting both sides by | |

11:11 | X will get that zero is equal to x squared | |

11:14 | minus x . And then if we factor out the | |

11:17 | G C F , if we factor our X , | |

11:20 | we'll get this expression Using the zero product property , | |

11:24 | We can see that X is equal to zero And | |

11:28 | X is also equal to one . So those are | |

11:31 | the points of intersection . Now the top function is | |

11:41 | our F of X function , which is Y equals | |

11:44 | X . The bottom function is R G of x | |

11:47 | function , which is x squared . So now we | |

11:52 | can calculate the area between the two curves . Using | |

11:55 | this formula , it's a definite integral from A to | |

11:58 | B of the top function F of x minus the | |

12:02 | bottom function G fx . Yeah , So this is | |

12:09 | going to be the integral from 0 to 1 . | |

12:11 | F five axes x Kiev axes x squared the anti | |

12:17 | derivative of X is going to be x squared over | |

12:20 | two . And for X squared using the power role | |

12:23 | is X to the third over three . So let's | |

12:33 | plug in one . This is going to be one | |

12:35 | squared over two minus one to the third over three | |

12:40 | . Then if we plug in zero , this whole | |

12:43 | thing will be zero . So it's 1/2 -1/3 . | |

12:50 | Let's multiply one half by three , every three and | |

12:53 | one of the three by to over to to get | |

12:56 | common denominators . So this becomes 3/6 . This becomes | |

13:02 | to over six And a final answer is one of | |

13:07 | the six . So this is the area between the | |

13:10 | two curves . It's one of the six square units | |

13:14 | . Number three , calculate the area of the region | |

13:18 | bounded by the curves , Y equals X squared and | |

13:21 | X equals y squared . So let's graph these two | |

13:26 | functions separately . So this is why is equal to | |

13:31 | X squared . It's a problem that opens upward now | |

13:37 | X equals y squared . It opens towards the right | |

13:46 | . If you take the square root of both sides | |

13:50 | , you will get that . Why is equal to | |

13:51 | square root of X cluster minus the top part of | |

13:55 | this function is why is equal to positive route X | |

14:05 | . The bottom part is why is equal to negative | |

14:07 | square root X . Now , if we focus on | |

14:17 | quadrant one where the two curves meet , this is | |

14:27 | why is equal to X squared and this is X | |

14:31 | is equal to y squared . So now this here | |

14:39 | is the top function which is F . Of X | |

14:45 | , which is basically that function X is equal to | |

14:47 | Y squared . But solving for Y , we know | |

14:51 | that it's the square root of X is equal to | |

14:53 | Y and F of X is equal to Y . | |

14:56 | And remember F f y is equal to X . | |

14:59 | So be careful . Don't get those confused . So | |

15:08 | we can replace F of X with the stuff that | |

15:11 | equals why ? Which is the square root of X | |

15:17 | ? The bottom function is why is equal to X | |

15:21 | squared . And so that's the idea of X . | |

15:23 | Function G of X is also equal to Y . | |

15:27 | Just a different type of way . So this is | |

15:29 | X squared . And our goal is to calculate the | |

15:33 | area of the shaded region . So first we need | |

15:40 | to find the point of intersection . So let's set | |

15:47 | F of X equal to G . Fx . So | |

15:51 | F of X . In terms of X is the | |

15:53 | square root of X . G of X is X | |

15:56 | squared . If we square both sides we'll get that | |

16:01 | X is equal to exit the fourth . Now this | |

16:06 | is true when X is zero , zero is equal | |

16:09 | to zero to the fourth . And this is true | |

16:11 | when X is equal to one , One is equal | |

16:13 | to 1 to the 4th power . So those the | |

16:16 | points of intersection . Yeah . Now that we have | |

16:20 | that we can calculate the area using this formula . | |

16:32 | So the area is gonna be the definite integral of | |

16:34 | a to B or 0-1 of the top function , | |

16:37 | which is the square root of X . Will write | |

16:40 | that as X to one half minus the bottom function | |

16:44 | , which is X squared . So the anti derivative | |

16:54 | of X to one half , we need to add | |

16:56 | 1 to 1 half Which History of the two . | |

17:00 | And then instead of dividing by three or 2 , | |

17:02 | we're gonna multiply by the reciprocal to over three . | |

17:05 | The anti derivative of X squared is X cube over | |

17:08 | three . Now Let's plug in one . So this | |

17:13 | is going to be to over three One of the | |

17:16 | three . And when we plug in zero we'll just | |

17:19 | get zero . So the final answer is just one | |

17:22 | of the three . So that's the area between . | |

17:27 | It's the area the region bounded by these two curves | |

17:30 | . So that's the final answer . Now let's move | |

17:32 | on to number four , calculate the area of the | |

17:35 | region bounded by the curves . X is equal to | |

17:38 | one minus five squared and X is equal to y | |

17:41 | squared minus one . So first let's talk about how | |

17:45 | we can graph this . We know that X is | |

17:49 | equal to y squared , looks like this . If | |

17:54 | that's X equal y squared what is exe equal Y | |

18:00 | Squared -1 . All we need to do is take | |

18:04 | this curve and subtracted by one along the X axis | |

18:09 | . So it's going to move one unit to the | |

18:10 | left , It's going to start at x equals negative | |

18:14 | one . So that's the shape of this graph , | |

18:21 | It shifted horizontally one units or less . So now | |

18:26 | that we have the graph of that function , let's | |

18:28 | focus on the graph of this function . Now X | |

18:33 | is equal to X equals positive . Wide square opens | |

18:36 | to the right . What about X equal negative Y | |

18:40 | squared . What happens then ? So this is going | |

18:44 | to open to the left , It's reflected about the | |

18:50 | y axis . Now what if we add a one | |

18:54 | to it When we subtracted by 1 ? The graph | |

18:58 | started at -1 . If we add one is going | |

19:01 | to start a positive one , So x equals negative | |

19:07 | y squared plus one . We're going to move one | |

19:10 | unit to the right , so this is the same | |

19:15 | as X is equal to one minus Y squared . | |

19:19 | So what we need to do is combine these two | |

19:22 | graphs together so it's going to look something like this | |

19:29 | and this is going to be the area of the | |

19:33 | region bounded by those two curves . So let's redraw | |

19:38 | that picture . So here we have why squared minus | |

19:53 | one . And here we have actually , this was | |

20:02 | one minus y squared and this one is going to | |

20:04 | be y squared minus one . So let me write | |

20:11 | that . So this is X is equal to Y | |

20:14 | Squared -1 . And this other one here is acts | |

20:17 | is equal to one minus y squared . So how | |

20:22 | can we determine the area of the region bounded by | |

20:26 | these two curves ? We need to find the points | |

20:32 | of intersection first . So let's set these two expressions | |

20:37 | equal to each other . And since we have acts | |

20:40 | in terms of why the point of intersection will be | |

20:44 | Y values . So let's set one minus y squared | |

20:48 | equal to y squared minus one . So let's add | |

20:54 | why square to both sides unless add one to both | |

20:58 | sides . So those will cancel , we'll get to | |

21:03 | is equal to two Y squared , Dividing both sides | |

21:07 | by two . We get one is equal to y | |

21:09 | squared And then if we take the square root of | |

21:12 | both sides we get one is equal to plus or | |

21:15 | minus one , which should be written this way . | |

21:23 | So this is a Y value of negative one and | |

21:26 | this is a Y value of positive one . So | |

21:33 | now that we have the y values of the points | |

21:36 | of intersection , we can calculate the area between the | |

21:39 | two curves . Using this formula , it's going to | |

21:42 | be the integral from C . Diddy , which are | |

21:45 | why values of the right function F of y minus | |

21:50 | the left function G f Y . So F of | |

21:54 | Y is here S S Y is one minus Y | |

22:02 | squared . We could see it here . It's the | |

22:05 | function on the right . G f Y is the | |

22:12 | function on the left . This is G F Y | |

22:16 | . It's on the left side for the region that's | |

22:18 | bounded , so G of y is Y Squared -1 | |

22:27 | . So now we could use this formula , it's | |

22:28 | going to be the integral from C to D C | |

22:32 | is -1 . He is positive one and then F | |

22:37 | of Y , which is one minus Y squared minus | |

22:42 | G F Y , which is Y Squared -1 . | |

22:49 | Dy . So let's simplify this expression first this is | |

22:56 | one minus y squared minus y squared . And then | |

23:00 | these two negative signs Will become positive one , so | |

23:09 | one plus one is two , negative Y squared minus | |

23:14 | Y squared is negative two , Y squared The anti | |

23:28 | derivative of two is too wide and for two I | |

23:31 | squared we have to I to the third over three | |

23:34 | Evaluated from -1 : one . So let's plug in | |

23:41 | one . This is going to be two times one | |

23:43 | -2 times wanted at 3rd over three . And then | |

23:48 | if we plug a negative one , well get this | |

23:57 | . So this becomes to minus two or three and | |

24:02 | then negative too Plus two of the three , Distributing | |

24:10 | the negative sign will have plus two -2 of its | |

24:14 | me . That was combine like terms two plus two | |

24:17 | is four -2 or 3 -2 or three . That's | |

24:22 | -4 or three . So what we need to do | |

24:27 | now is we need to get common two nanometers . | |

24:32 | So four of the one . I'm going to multiply | |

24:34 | that by three of us three . So this becomes | |

24:40 | 12/3 -4 or three , Which gives me a final | |

24:45 | answer of 8/3 . So this here is the area | |

24:49 | between the two curves . Now , let's work on | |

24:52 | some more practice problems calculate the area of the region | |

24:56 | bounded by the line Y equals x squared minus four | |

24:59 | X . And the X axis X . Where we | |

25:07 | know opens like this X squared minus four X . | |

25:15 | It's going to be similar but it's going to be | |

25:18 | shifted . So what we need to do is to | |

25:23 | graph this , we need to find the X intercepts | |

25:28 | and the vertex , So the final x intercepts . | |

25:34 | Let's replace why ? With 0 ? Let's solve for | |

25:36 | X . So I'm going to factor on X . | |

25:43 | Using the zero product property , We get an x | |

25:46 | intercept of zero and of four . So let's make | |

25:52 | a table . So when X zero , Y is | |

25:57 | zero When X is four , Y is also zero | |

26:03 | . Now the vertex of a parabola Is the midpoint | |

26:10 | of the x intercepts . So it's going to be | |

26:12 | at two . Another way you can in which you | |

26:15 | can get the vertex Is by using this equation X | |

26:19 | . is equal to negative B over two . A | |

26:23 | . Let's see if you have a quadratic equation A | |

26:26 | . X squared plus bx plus C . So is | |

26:30 | the number in front of X squared which is one | |

26:34 | . B . Is the number of front of X | |

26:36 | Which is negative for . So we have negative B | |

26:39 | over two . A . So this is positive for | |

26:41 | over two , which gives us an x . value | |

26:44 | of two . Now we need to determine the Y | |

26:47 | value to do that . We can just plug it | |

26:50 | into this expression . So why is equal to X | |

26:54 | times x minus four ? If we plug into this | |

26:57 | is two times two minus four . So we get | |

27:01 | two times negative two . Which is negative for . | |

27:05 | So we have enough information . Oops , let's go | |

27:11 | back . So we have enough information to graph this | |

27:20 | function out . So we have an exodus up of | |

27:35 | zero and 4 And that -4 . We have the | |

27:47 | vertex at an x . value of two . So | |

27:52 | the graph is a problem that opens upward like this | |

28:00 | . Our goal is to calculate the area of the | |

28:02 | region bounded by that line . Well technically that should | |

28:06 | be a curve nothing like . So let's just replace | |

28:11 | that with the word curve and the X axis . | |

28:16 | So how can we find the area of that shade | |

28:18 | of region ? Well , we need to identify the | |

28:23 | top function and the bottom function . The top function | |

28:27 | is the X axis . And that's basically the line | |

28:32 | of y equals zero . So we could say the | |

28:35 | top function is F . Of X Is equal to | |

28:39 | zero . The bottom function G fx is the curve | |

28:44 | , X squared minus four X . And we're gonna | |

28:54 | integrate this from zero . It's A four . So | |

28:58 | let's use this formula , it's going to be the | |

29:01 | integral from A to B . Of the top function | |

29:05 | F of x minus the bottom function G fx . | |

29:14 | So this is gonna be the definite integral from 0-4 | |

29:17 | . The top function is zero . The bottom function | |

29:20 | is negative . Well it's minus X squared minus four | |

29:25 | X . So let's go ahead and distribute the negative | |

29:29 | sign . So negative times negative four X . That's | |

29:36 | going to be positive for X and then minus X | |

29:40 | squared mm The anti derivative of four X is going | |

29:49 | to be four X squared over two and for X | |

29:52 | squared is going to be X cube over three . | |

30:04 | So this becomes two X squared . So we have | |

30:10 | two X squared minus excuse over three , Evaluated from | |

30:20 | 0 to 4 . So let's plug in four . | |

30:31 | When we plug in zero , this whole thing is | |

30:33 | going to be zero , Four squared is 16 times | |

30:38 | two . That's 32 . 4 to the 3rd is | |

30:41 | 64 . So we have 64 3 . Now we | |

30:45 | need to get common denominators . Let's multiply 32 of | |

30:49 | the one By three of the three , 32 Times | |

31:03 | Street is 96 . So we have 96/3 -64 or | |

31:07 | three And 96 -64 is 32 . So the area | |

31:14 | of the shade a region Is 32 of the three | |

31:18 | . That's the answer for this problem . Number six | |

31:23 | , calculate the area of the region bounded by the | |

31:25 | equations , Y equals X squared minus four X and | |

31:29 | y equals six minus three X . So this equation | |

31:39 | we're familiar with already , We know the x intercepts | |

31:43 | will be zero and 4 and we know that the | |

31:51 | vertex Will be at 2 -4 . So this graph | |

31:59 | is going to open upward like this . That's a | |

32:05 | rough sketch . Now . Here we have a linear | |

32:09 | equation . Y equals six minus three X . So | |

32:12 | the Y intercept is six , which should be somewhere | |

32:22 | here . And to determine the exodus up . We | |

32:26 | can replace why with zero and solve for X . | |

32:29 | So adding three X to both sides . We have | |

32:31 | three acts is equal to six , and six divided | |

32:35 | by three is too . So this line is going | |

32:39 | to touch the X axis . That too . So | |

32:48 | that's what we have . And eventually this curve is | |

32:52 | going to meet up with this line . So our | |

32:57 | goal is to get the area of the shaded region | |

33:06 | . Well , we need to determine is the point | |

33:09 | of intersection here and here . So we need to | |

33:16 | calculate the area by taking the top function and subtracting | |

33:20 | by the bottom function . So we want the X | |

33:23 | values of the points of intersection . So let's set | |

33:35 | these two functions equal to each other X squared minus | |

33:40 | four X . We're going to set that equal to | |

33:41 | six . The ministry acts . So let's begin by | |

33:50 | let's add reacts to both sides Unless attract both sides | |

33:55 | by six . So on the left side will have | |

34:04 | X squared -4 . X plus reacts . That's going | |

34:08 | to be negative X . And then we have negative | |

34:11 | six . So we could factor X squared minus x | |

34:18 | minus six . Two . Numbers that multiply to negative | |

34:22 | six . But that's the middle coefficient negative one are | |

34:25 | negative three and plus two . So we can write | |

34:28 | this as X -3 times x plus two . Some | |

34:33 | for X . We get that X is equal to | |

34:35 | positive three . And if you set X plus two | |

34:38 | equal to zero , you'll get that X is equal | |

34:40 | to negative two . So here's the first point of | |

34:45 | intersection . This is that an x . value of | |

34:48 | three And this is at an x . value of | |

34:51 | -2 . And the top function that's F . Of | |

35:00 | X . Which is it's this linear equation 6 -3 | |

35:07 | x . The bottom function G fx Is the other | |

35:11 | one that's X squared minus four X . So now | |

35:20 | let's calculate the area using this formula . So it's | |

35:34 | going to be the integral from A to B . | |

35:36 | That is from negative 2 to 3 F . Of | |

35:39 | X . Which is 6 -3 . X minus G | |

35:45 | . Fx which is X squared minus four X . | |

36:00 | So we have six . This is negative three , | |

36:03 | X minus negative four X . Which is negative three | |

36:06 | X plus four X . So that's simply plus X | |

36:11 | . And then we have negative X squared . Now | |

36:20 | let's clear away a few things . The anti derivative | |

36:32 | of six is going to be six x for X | |

36:36 | . Is going to be X squared over two and | |

36:39 | for X squared is going to be exited third over | |

36:41 | three , Evaluated from -2 - three . Now Let's | |

36:49 | plug in three . So we're gonna have six times | |

36:52 | street plus three squared over to - Streets . The | |

36:58 | third power over three and that was fucking negative too | |

37:17 | . six times 3 is 18 . three Squared is | |

37:20 | nine . 3 to the 3rd is 27 divided by | |

37:24 | three . That's 9 . And then we have negative | |

37:28 | six times negative two is negative 12 negative two squared | |

37:32 | is four , divided by two . Is positive too | |

37:35 | -2 to the third . Power is negative eight With | |

37:39 | the negative front that becomes positive eight but divided by | |

37:41 | three . So here we can combine eight . I | |

37:47 | mean 18 and negative nine , 18 -9 is nine | |

37:55 | . And here we have negative 12 plus two , | |

38:00 | Which is -10 . And then plus 8/3 , Distributing | |

38:07 | the negative sign . We have nine Plus nine of | |

38:10 | it too . And then we have plus 10 -8/3 | |

38:22 | . So first let's combine nine and 10 . So | |

38:26 | that's just 19 . And then let's focus on combining | |

38:31 | those two fractions . Let's try to get a common | |

38:42 | denominator of six . So 19/1 . I'm going to | |

38:47 | multiply that by 6/6 , 9/2 . I'm going to | |

38:51 | multiply that by three of the three and 8/3 . | |

38:55 | I'm going to multiply that by 2/2 , 19 times | |

39:04 | six Is 114 . nine Times Street is 27 , | |

39:12 | eight times 2 is 16 , 27 -16 is 11 | |

39:23 | and 1 14 plus 11 is 1 25 . So | |

39:27 | this is the final answer . This is the area | |

39:29 | of the shaded region . It's 125/6 , # seven | |

39:37 | . So we want to find the area of the | |

39:38 | region bounded by these two equations . We got a | |

39:41 | linear equation in a parabolic equation . Let's begin by | |

39:46 | graph in the parabolic equation . So we have X | |

39:50 | equals y squared , which we know it's going to | |

39:53 | open towards the right , but it's going to start | |

39:55 | at negative four along the X axis . So it's | |

40:05 | gonna look something like this . Now , here we | |

40:11 | have a linear equation , X is equal to three | |

40:14 | Y -2 . So when why is 0 ? X | |

40:20 | will be negative too . So that means that the | |

40:26 | x intercept is -2 . Now we have a positive | |

40:31 | coefficient in front of why ? It's positive three white | |

40:34 | and that negative three Y . So we know we're | |

40:37 | going to have a linear equation that's going to increase | |

40:42 | as X increases as Y increases , X . Is | |

40:49 | going to increase . So it's going to go up | |

40:54 | . So we want to find the area of the | |

40:56 | shaded region . What we need to do is find | |

41:00 | the points of intersection first . So let's set three | |

41:05 | Y -2 equal 22 Y squared minus four . Some | |

41:12 | get to take these numbers and moved to the other | |

41:15 | side . So we're gonna have zero is equal to | |

41:18 | two Y squared . This will be negative three Y | |

41:22 | on the right side And then this will be positive | |

41:25 | two on the right , so negative four plus positive | |

41:29 | two . That's gonna be negative too . So how | |

41:32 | can we factor this ? Try no meal what the | |

41:34 | leading coefficients to to factor it . We need to | |

41:38 | multiply the length coefficient by the constant term two times | |

41:42 | negative two is negative for . And then we need | |

41:46 | to find two numbers that multiply the negative four but | |

41:48 | add to the middle coefficient negative three . This is | |

41:51 | going to be negative for and plus one . So | |

41:54 | we're going to replace the middle term negative three . | |

41:57 | Y . With -4 Y Plus one , Y . | |

42:03 | And then we could factor by grouping . So we're | |

42:05 | gonna take out the G C F in the first | |

42:07 | two terms and that's too wide . It will be | |

42:10 | less with why minus negative too . We're going to | |

42:14 | take out the G C . F in the last | |

42:15 | two terms , Which is just gonna be one And | |

42:18 | then times why -2 . Now these two are the | |

42:21 | same . So we're going to factor out why minus | |

42:25 | two And we're left with two I plus one . | |

42:33 | So that's how we can factor this . Binomial if | |

42:37 | we set why mine is to equal to zero we'll | |

42:39 | get y . Is equal to positive too . And | |

42:43 | if we set to Y plus one equal to zero | |

42:46 | , Why it's going to be a negative 1/2 . | |

42:50 | So now we have our points of intersection And these | |

42:54 | are why values negative 1/2 and positive too . So | |

42:58 | this is equal to see and this is equal to | |

43:01 | D . Since everything is in terms of why here | |

43:12 | . So now we could use this formula to get | |

43:14 | the area . It's going to be the integral from | |

43:17 | C . Diddy of the function on the right , | |

43:19 | which is F A Y minus the function on the | |

43:22 | less which is G F . Y . So relative | |

43:27 | to the shaded region , the function on the right | |

43:31 | , which is here . That function is S of | |

43:36 | Hawaii which is the linear equation three , Y minus | |

43:42 | two . The function on the left , that's the | |

43:46 | curve function , that's G . Of Y . Which | |

43:49 | has to be the other one too , I sq | |

43:51 | -4 . So the area is going to be from | |

43:55 | C . To D . Or from negative one half | |

43:58 | . It's it's you F . of Y . Which | |

44:00 | is three Y -2 . And then by she of | |

44:04 | Y , which is two Y squared minus four . | |

44:08 | Dy So we have three Y -2 minus two Y | |

44:20 | squared And then plus four . So this becomes negative | |

44:32 | two Y squared plus three Y . And then combining | |

44:36 | east to That's going to be plus two . The | |

44:55 | anti derivative of negative two Y squared will be negative | |

44:59 | two Y to the third over three . For three | |

45:02 | Y . It's three Y squared up to And for | |

45:05 | two it's too high evaluated from -1/2 to 2 . | |

45:15 | So now let's plug into this is going to be | |

45:19 | negative too . Times 2 to the third , power | |

45:22 | over three plus three times two squared over two Plus | |

45:29 | two times 2 . And then let's fuckin negative 1/2 | |

45:34 | . So it's negative two times negative 1/2 To the | |

45:38 | third . Power over three plus three times negative one | |

45:43 | half squared over to plus two times -1 . Two | |

45:51 | to the third . Powers eight times negative two . | |

45:54 | That's negative . 16/3 . Two squared is four divided | |

45:59 | by two , is two times three that gives us | |

46:03 | six And then two times 2 is four negative one | |

46:10 | half to the third . Power is negative 1/8 , | |

46:14 | negative 1/8 times negative two is positive . One of | |

46:18 | the four Divided by three that becomes positive . One | |

46:21 | of the 12 negative one half squared is positive 1/4 | |

46:28 | Divided by two . That's one of her eight Time | |

46:30 | Street . That's positive three or 8 . And then | |

46:37 | two times negative a half is negative one . So | |

46:43 | we have negative 16/3 , 6-plus 4 is 10 , | |

46:48 | distributing a negative sign . We have negative one of | |

46:50 | the 12 -3 of eight . And then this is | |

46:54 | plus one . So what we need to do is | |

47:04 | we need to get a common denominator Of 3 , | |

47:07 | 12 and eight . The best number is 24 . | |

47:11 | 24 is a multiple of 3 , 12 and eight | |

47:15 | . But first Let's combine 10 and one , which | |

47:19 | is 11 . We're going to multiply this by 24/24 | |

47:28 | and then negative 16/3 . We're going to multiply that | |

47:31 | by 8/8 , -1 of the 12 will multiply that | |

47:35 | by two x 2 . A negative 3/8 . We're | |

47:39 | gonna multiply that by 3/3 To get a common denominator | |

47:43 | of 24 . So we have 11 Times 24 , | |

47:52 | which is to 64 16 times eight is 1 28 | |

48:02 | , two times 1 is two , three times 3 | |

48:05 | is nine , So to 64 -128 , That's 136 | |

48:15 | and -2 -9 is negative . 11 1 36 -11 | |

48:22 | Is 125 . So the answer is 125 over 24 | |

48:28 | square units , so that is the area Of the | |

48:31 | region bounded by the two curves or the two equations | |

00:0-1 | . |

#### DESCRIPTION:

This calculus video tutorial provides a basic introduction in finding the area between two curves with respect to y and with respect to x. It explains how to set up the definite integral to calculate the area of the shaded region bounded by the two curves. In order to find the points of intersection, you need to set the two curves equal to each other and solve for x or y. You need to be familiar with some basic integration techniques for this lesson. This video contains plenty of examples and practice problems.

#### OVERVIEW:

Area Between Two Curves is a free educational video by The Organic Chemistry Tutor.

This page not only allows students and teachers view Area Between Two Curves videos but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.

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