Area Between Two Curves - Free Educational videos for Students in K-12

Area Between Two Curves - By The Organic Chemistry Tutor

Transcript


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	.

Summarizer

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.

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