BAM! Mr. Tarrou. In this lesson we are going to extend our understanding of the first fundamental theorem of calculus and apply it to finding

the area between 2 curves. We are going to introduce the basics of these types of problems,

we are going to go through 4 examples, which I feel go from easier to more difficult, the

last one is going to involve a little bit of trigonometry and some understanding of

odd functions, which will hopefully allow us to kind of shorten the process of finding

the final answer. All these problems have been written to be solvable without the aid

of a calculator but if you are given the problems that do require the use of calculators, we

will throw a little video in the middle of helping you understand how to do that with

a calculator, finding the intersection of 2 graphs and doing a numerical integration

with your calculator. Okay, so the four examples, you will find time stamps in the description

of this video if you want to skip ahead and just get to those examples but for the basic

introduction, let me get out of the way because I did use up the whole board up here to make

sure I can fit my 3 diagrams. If f and g are continuous on a closed interval a, b. And

if g(x) is less than f(x) for all x in the closed interval, then the area of the region

bounded by the graphs of f and g and the vertical lines of x=a and x=b is, our area is equal

to the definite integral from a to b of [f(x)-g(x)] dx, if we are integrating with respect to

x, cause that’s our independent variable the way this formula is written. So what I

have up here is 3 diagrams to illustrate to you that it doesn’t matter if the functions

are above or below of the x axis, we should have understood from our work with application

problems maybe, working with the first fundamental theorem of calculus that if the function falls

below the x axis then all of those sums are going to be negative. And we can’t have

a negative area. SO, you might go, okay well, I have a closed interval from a to b and we

have our function f(x) is always greater than g(x) like our theorem here requires and we

are going to, just like we were first introduced to finding, estimating area with a finite

number of rectangles, which then moved into limit process which then moved into the first

fundamental theorem of calculus. We have a little representative rectangle here, which

we would be using an infinite number of these to estimate the area and the height is found

by doing f(x)-g(x) and the width is dx or delta x, well these are both above the x axis,

so it would kind of with our understanding of the first fundamental theorem of calculus,

say, if I found the area bound between the function f(x) and the x axis on the closed

interval of a to b, I would have all this area shaded and all going down to the x axis.

So then if I just found the area bound between g(x) and the x axis, I would just subtract

that which kind of looks exactly like this theorem says or this rule says for us to find

the area bound between two curves. Well now we have g(x) is below the x axis and over

here I have got g(x) and have pushed them down so that both function f and function

g are below the x axis. And then we only have one curve and again if that curve falls below

the x axis, we understood that the sums became negative which was okay for our application

problems like velocity where we have a positive-negative movement…. but not for area. Remember we

are just finding the area of a bunch of rectangles, which is width times height, so as long as

the width and the height are positive, of course we are going to have a positive area.

And I am kind of just making up some numbers here. Let’s just say, for instance, this

function f and this function g at c, that our y values, the values coming out of our

functions are 8 and 3. Well that’s going to give this particular rectangle at a value

of c, a height of 5, and that would be some kind of width times height, and that would

give us an estimate of the area of that rectangle. Well if I push both of those functions down

by 4, and again, I almost wrote this is as f(2) and g(2), I just want you to show that

I am pushing these down 4 units and so the distance between these functions would stay

the same. Well 8-4=4 and 3-4=-1. And still doing that f(x)-g(x) idea, sort of, for a

concrete example, we have 4-1, that y value of 4 from f(x) and that that idea of this

y value of -1 from g(x), well 4- (-1) is still going to give us a height of 5, so that rectangle

would still have the same width, and at least at the value of c, trying to match all these

graphs up, we have a height of 5 and thus we can see that the area would still be positive

between these 2 graphs as long as f(x) is always greater than g(x). And if I pushed

them both below the x axis, with a vertical shift of down 10, on both of the original

functions of f(x) and g(x), well 8-10 is -2 and 3-10=-7. And -2 minus -7 is still that

height for that rectangle, giving us the same value of 5. So as you find these areas between

2 curves, again, the restriction here is that it be continuous on a closed interval, f(x)

must be greater than g(x) on that entire closed interval and if it is, the area is again the

definite integral from a to b of [f(x)-g(x)] dx and it doesn’t matter if one or both

of those functions are below the x axis. BAM! Now the previous page said that our area,

which we were finding between 2 curves, was bound by the vertical line of x=a and x=b

which is requirement if the functions never intersect, it can still be requirement if

the functions intersect but a lot of functions simply say, find the area between the intersecting

curves, and thus you have to find where those 2 curves intersect, because they are going

to define the lower and upper limits of our definite integral, so, guess what you have

to do, find the intersection points, that will either be algebraically or with the aid

of a calculator depending on the difficulty of the problem and the requirements that your

teacher is asking of you. And then the area is going to be definite integral from a to

b of [f(x)-g(x)] dx Now let me talk a little bit here about why are the x values of my

2 intersection points defining the lower and upper limits of my definite integral because

that’s not always going to be the case, usually but not always. If you look at this notation,

we are taking an x in and plugging it into a function f and finding f(x) or finding a

y. So it’s the y values of the function that are defining the height of the rectangle and

then therefore the width of the rectangle is defined as dx or delta x. so we are using

the x value, horizontally, right, the values along this horizontal axis to define the lower

and upper limits of our definite integral because the y values of the function, if you

will, are allowing us to find the height of the rectangle so if you want to find the area

and let the x values determine the lower and upper limit, you are going to be doing that

when you are talking about functions which are in terms of x and those functions giving

us the y values which we are subtracting or you could say, if you are using vertical rectangles,

then you are finding area along the horizontal distance to find a value or that’s defining

our closed interval. Okay if we have functions which intersect more than just twice, thus

they are going to set up multiple areas which we have to find the area of and add up all

those pieces, well in this graph, we have function g(x) is greater than function f(x)

between the intersection points of (a,e) and (b,f), so, okay we will find the area of that

part or that portion, or that area bound by that interesting graph by doing area is equal

to the definite integral from a to b of g(x)-f(x) dx Again using those vertical rectangles and

having the widths defined as delta x because we are integrating with respect to x. But

after to the right of this intersection point, the function f(x) now becomes greater than

function g(x). SO we’re going to have to find this area where it’s g(x)-f(x) and find the

area bound between these intersecting curves which is going to be f(x)-g(x) since f (x)

is greater and that would be the area of the first region plus the area found by doing

the, Excuse me, definite integral from b to c of f(x) -g(x) dx. Now this third diagram,

this is going to be one of those difficult problems we get to as an example, I have got,

just to have something to look at and explain why it is going to be difficult to find the

area using a bunch of vertical rectangles. We have a sideways parabola, if you will,

g(x)=plus or minus the square root of (x-1) +3 and then another unknown function here

of f(x). If you are going to find this area, with a bunch of infinite vertical rectangles,

we have an intersection point here, let’s just go ahead and call this point (b,e) And

now we have another intersection point over here which we are going to call (c,f) And

the previous page and these first 2 examples, we found the function that was on top, the

greater than function and subtracted it with the function that was lower or less than,

right? Well, if come down here and draw a vertical line and the vertical line is going

to be x=b, then this side is pretty obvious, function f is greater than function g(x),

now it is plus or minus the square root of (x-1)+3, so that means that we have some,

what we would, with the parabola, be calling this the vertex, and we label this as point

(a,g) Well then we are talking about the lower half , so this is going to be something like

f(x)- negative square root of (x-1)+3, but then what do you do over here, because, it’s

sort of like, this parabola is going to be its own upper and lower values, so you are

going to have to account for a horizontal line going over here and we would be and subtracting

this function which is the square root of (x-1)+3 and subtracting it with this part

of the function which is negative square root of (x-1)+3. So let me get out of the way,

erase this board and show what it would look like, I am not giving you function f, because

we are going to do that in great detail with the third example, I think that’s what we

are going to do, but let me just clear this off and show you a little bit of an idea of

what this would look like algebraically to find that area if you are going to use vertical

rectangles and integrate with respect to x. So, if again, this is a relatively simple

example with this just being a sideways opening parabola, I am able, or with this concrete

example, to find this vertex and if you know the vertex form or the standard form of the

parabolas, then this would be the x coordinate of the vertex which is 1 and the y coordinate

=3 so this point is (1, 3). Now, again, we would find this area here, well the area is

equal to the definite integral from 1, that’s the x value of this vertex going to the upper

limit of b, and this area would be found by, well this y value is square root of (x-1)

+3 minus these y values down here which is negative square root of (x-1) +3. And so this

first part of the equation would find this area and then the second part, like I said,

that’s a bit more straight forward, that would be the definite integral from b to c, what

I have labelled as my two x values here, my intersection points of f(x), okay the way

this is drawn, f(x) is above my bottom half of the parabola, so minus negative square

root of (x-1) +3. Well that’s rather complicated and finding the integrals may possibly be

not very simple, so if I would rather use, let’s see another color piece of chalk,

if I would rather use horizontal rectangles, then instead of having the widths defined

as dx, my widths would be defined as dy. Now if I am using horizontal rectangles, with

a width of dy or delta y, then it’s going to be the y values that define my lower and

upper limits, so you need to turn these equations around and not have them in terms of x but

have them in terms of y, so that would mean for this one, I would subtract both sides

by 3, then I would square both sides of the equation and, I have started to run out of

room, I would then add 1 to both sides and then write that function in terms of y we

would have, some kind of x function now in terms of y and this would have to be rewritten

also to be in terms of y, so I am just going to write this as f sub 2 (y), so I don’t

know what the function is, but turn the values around and solve for x and have some kind

of function in terms of y. Now that would be a lot less work, because I saw the intersection

points which we would have to find first, but as you read, where I have my large, for

any particular value of y, the larger x value so I have the larger x’s, right..because when

in plug in the y into the function, I would be getting an x. so I have these larger x’s

and then I have these smaller x’s. So I will be doing this function which is f(x) minus

this function over here with the smaller x’s for each respective value of y we’ll be plugging

it into the function. And we would have a simpler definite integral for us to find the

area between those 2 intersecting curves. That would be the area=the definite integral

the lower y value here is f and the upper one is e, so it would be the definite integral

from f to e of, again that reading up from right to left, or larger x’s and then smaller

x’s, f sub 2 of y, minus, maybe I should call this g sub 2 of y , but I gave us a concrete

equation for that one, so it would be minus (y-3)^2 +1 and again we are integrating with

respect to y. So it’s this kind of function here that is we do our third example, we are

not going to want to find the area with a bunch of vertical rectangles because it’s

going to really complicate the process, we are going to find it with horizontal ones

instead and have some simpler work, as long as you can write these functions that are

given to you, turn them around and have some sort of x in terms of y, you can do that easier.

So let’s go on to the first example, I think. NO! I am going to give you some sort of formal

language on how to do these rectangles horizontally. Putting into a formal statement, what we did

in the former, sort of the end of the previous ‘non -example’ example, horizontal representative

rectangles, , cause I didn’t really give you those functions, reversing the rules of

x any y, If w and v are continuous functions and if w(y) is greater than or equal to v(y)

for all y in closed interval [c,d], then the area of the region bound on the left by x=v(y)

and just looking to make sure, I have this drawn correctly, and on the right by x=w(y)

below by y=c and above by y=d, then the area is found by the definite integral from c to

d of w(y) -v(y) with respect to y or dy. So again we have, it doesn’t have to be always

a sideways parabola, but we have a function v, a function w(y) and that’s just the diagram

that goes along with this statement. Again using those horizontal representative rectangles,

and we know that we have horizontal representative rectangles if we are integrating with respect

to y. Okay, now before we move on to our concrete examples with all those numbers in them and

real functions, this particular example looks kind of like the other one and whether you

find it easier to find the area bound by actually 3 functions, here with vertical representative

rectangles or horizontal ones, it’s going to depend on the functions, because if you

want to use horizontal rectangles, then there’s still a dividing line here, well we’re going

to be doing, these are in terms of x, so we are going to have to turn those around, but

it would be taking the y values or the what will it be, it will be f sub 2 of y, taking

these values here and subtracting it with this function, for this lower bound and then

this orange graph or f sub 2 of y, if you will, minus g sub 2 of y. We had to turn both

of those functions around, but the point is that you would still have to set up two definite

integrals whether you were finding this area with horizontal representative rectangles

or vertical ones, because if you want to find the area bound by these functions by using

vertical representative rectangles which is how my notation is set up, of course, with

my independent variables of x, then you are still going to have to use 2 definite integrals.

So that would be, with the way this is drawn, the area=the definite integral from.., after

finding the intersection points, from c to a of, the greater function is g(x) minus h(x)

dx + the definite integral from e to c of, our greater function is f(x) minus, again

h(x). So unlike, the last sort of example, we started talking about let us use the horizontal

representative rectangles, some kind of diagram like this- horizontally or vertically you

still need to find the area using 2 definite integrals and again how we decide to do this

is going to be determined by how the functions are set up, as opposed to graphically or with

the number of definite integration expressions we need to set up. Alright, let’s go to

some actual examples. Sketch the region bounded by the graphs of the algebraic functions & find

the area of the region. The functions are y=-2/3 x times(x-6) and y=-1/2 x+11, x=0

and x=6. Why do we need to sketch the picture, the graph these functions are forming..? You

know it looks like they gave us a couple of vertical lines so seems like these should

be our lower and upper limits, but what function is above the other, like what is the greater

function? Do they intersect on this closed interval where we are going to find the area,

and thus requiring us to set up 2 integration expressions as opposed to just one? So let’s

get out of the way. You can pause this video and do this on your own. Let’s find out what

these functions are going to look like, that of course is obviously going to be a line

and allow that graph to help us set up the definite integral. So, we of course we did

the -2/3 x and distributed it through the parentheses. I noticed that it was a parabola,

which you can put into standard form or vertex form or whatever your textbook calls it or

I found the x coordinate of the vertex to be -b/2a. Getting the coordinate of the vertex

to be 3 and the y coordinate from that 3 of 6. And…I didn’t really have to find the

x intercepts really cause we already have our left boundary of x=0 and x=6, but the

parabola is opening down and they have the x intercepts going at 0 and 6. SO, this would

be, I think, much simpler, clearly, if we find the area using vertical representative

rectangles and thus our area is going to be found by the definite integral from 0 to 6

of, now we have written the rule, it said f(x) -g(x), oh by the way, I relabeled these

functions, cause didn’t want to have two y’s, have them same name for two different

functions in the same problem, that’s an issue. And the way I have labeled, I have invariably

put g(x) on top, it is the greater function which is perfectly fine. And that straight

line, that g(x) that was -1/2x+11 minus the other function our downward opening parabola,

of… I might as well use this one, -2/3x^2 +4x dx. SO, let’s get this out of the way,

hopefully by now we are very comfortable with evaluating definite integrals, I am going

to pull out and move this ”screen…”‘ out of the way and reveal the answer, one step

at a time if you want to practice your integration skills, just pause the video and try it on

your own first, before you know, I get to the answer. SO hopefully, you found that one

to be pretty straight forward, negative times a negative is positive 2/3 x^2, negative times

4x is negative 4x. -4 +(-1/2) comes out to be a -9/2x +11, this is just a basic polynomial

so raising that power by one and dividing, we get 2/9; raising that power by 1 and dividing,

we get -9/4 x^2 and then finally 11x and we find the integral with respect to x. And then

finish up by putting in the..doing this with the x substituted in and minus our lower limit

of zero, then we get an area of 33 square units. Okay next example is going to involve

a couple of slightly more complex polynomials and they are going to intersect couple of

times or few times actually setting up 2 areas which we’ll need to find the individual areas

and add those results for the final answer. With the same directions, we have the functions

of -4x+y=12 and y=x^3 +3x^2. And we did not get any vertical or horizontal lines..bounding

the area. So, we are looking for the area formed by the intersection of these 2 graphs.

I am going to go ahead and sketch the cubic graph for us and that basically is just a

line 4x+12, hopefully you will already have this diagram given to you or you’ll have the

ability to use a calculator to help you graph that cubic function, otherwise you are going

to have to find x and y intercepts and remember all the steps we had to go through to graph

polynomial functions. At any rate, you do need to find the intersection points so we

know how we are going to set up the definite integrals. If you look at the way this functions

curves around, think about what you would have to do to using with horizontal representative

rectangles where, if you are doing it horizontally, the greater bound and lower bound are maybe

the same function and that is going to be very very difficult trying to figure out.

So this would be a much easier problem to do with vertical representative rectangles.

Again, we are going to find these intersection points and allow them to help us set up the

lower and upper limits of the definite integrals. SO grab some chalk here, we are going to set

these equations equal to each other and find those intercepts, now again I am trying to

make some problems that I can relatively easily do on camera, so this will be solvable without

the aid of a calculator, not stumble too much around for our example in our video. So we

have this equation. These 2 parts are equal so we are going to move everything to one

side and set it equal to zero. And solving these high order polynomials, well you could

be talking about having to find a long list of possible rational zeros, and then doing

synthetic division until you get a remainder of zero, and keep doing that process till

you get to second degree and then finish with a quadratic formula. If you have the ability

to use a calculator un your class, of course, graph this equation and find the zeros or

roots or x intercepts graphically with your calculator using some kind of calculation

command or graphical from menu. Or hopefully, that is as easy as it gets, if they are factorable

by grouping which I said..we can do this relatively cleanly. We can factor this by grouping. So

what comes out of these two terms? They both have a common factor of x^2. x^3 divided by

x^2 is x so we have x+3. We can take out a factor of -4 from these last two terms. And

now we have two terms that both have a common factor of (x+3). So we can take that out.

And then last step here, we still have a factor that’s factorable again and that’s the

difference of two perfect squares and that’s going to give us zeros of x=-3, x=-2 and then

x=2 and since we are going to find the area bound between these two functions using vertical

representative rectangles, this is all we need to set up the lower and upper limits

of those definite integrals. I am going to do that on screen and I am going to clear

this up, I will go ahead and give us the full coordinates of those points just to let us

know what they are. With room to work and our intersection points identified, we are

going to go ahead and set up that definite integrals. We have this first area so we are

going to find that area by setting up definite integral of… our lower limit is -3 and upper

limit is -2 and that will be of this function on top is g so (x^3 + 3x^2) and if you want

to wrap up the entire polynomial function to emphasize that substitution as a good idea

but it’s not going to do much expect that minus sign here. If we don’t distribute that

through the second function we are going to have sign errors and get the wrong answer.

And the first area lower function is f (x) so we have 4 x +12 and of course we are integrating

with respect to x. So don’t forget the dx. Now moving on to the second area we are going

to add the definite integral lower limit is – 2 upper limit is +2, our upper function

is f(x) now so 4x+12 minus our lower function is (x^3 +3x^2). Now these are pretty straight

forward polynomials and we have been doing integration for quite a while so I am going

to go ahead and move that out of the way and reveal this in small steps. You can again

pause the video and try on your own if you like but I am not going to go through every

single step slowly. BAM!! 131/4 square units after you take the area of these two regions

and add them up. Now clearly this was easily solved by hand and it was an easy polynomial

for us to do our integration with, just using the power rule. If you couldn’t find these

intersection points by hand or the function did not fall into.., or we couldn’t use

some of our fundamental integration rules to integrate these functions then you would

have to use a calculator to find these intersection points and find the area of these regions

so you could then add them up and get the answer. So, even though we can do this problem

by hand, I just want to, before we get on to the third example, throw in a little clip

of how to this with a calculator. CAS… and I want to do this in the same way so you could

do this on a TI-83 or a TI-84. We are going to go to ScratchPad and get the graphing up.

We already have the entry window open and our graph was 4x +12. Hit enter. Hit the tab

button again to open up that entry line and we have x^3, hit the arrow to get the exponent,

plus 3x^2. Enter. Alright. Well I can’t see the intersection of these 2 graphs, I

am going to go to Menu, go to Window/Zoom and do Window settings. We are just going

to close up that window a little bit, maybe -7 and x max of +7 and we went low enough.

Let’s set y max be 24. And there we go. Now we are going to find both of the areas.

If you understand one, you can do both. So let’s find the bigger area here. Now we’re

going to find the area below the linear function and between the intersection points of -2

and +2. And then we are going to do the same thing with the cubic function and then subtract

them. So first, we are going to need to find the intersection point, assuming we don’t

already have them with us. So we are going to go to Menu – Analyze graph- find the

intersection. The calculator will ask you for the lower and upper bound, I am using

my touch pad of course but you will be using this pad here on your calculator and there

our intersection of (-2, 4) And then finally again, …. Menu – Analyze graph , using

touchpad, find the intersection- Lower bound and upper bound and there is the other intersection

point of (2, 20). Alright so, I want to find the area below the linear function of 4x +12

between these values. Menu-Analyze graph – let’s find the integral. The calculator will ask

you what graph. We want to do this one, so we got to do this one graph at a time. Lower

bound – you can just type in the number so I am going to go -2 and there is a little

box opening up here. Enter and you will see that actually with, the INSPIRE will show

me the area as I move my upper bound around. But I want to get that set to a concrete value

of 2 so I am just going to hit 2. There is a little window opening up for me and we have

that area of 48. I am going to write that down on my piece of paper. Now we are going

to find the area below the cubic function. So again Menu- Analyze graph- integral- which

graph. Well I want to use the cubic one this time. Lower bound -2 I am using my key board

here, but you will be hitting negative here and enter. Lock that up. Upper bound, I want

to again have that to be a value of 2. So I will just hit 2 and enter and the area below

the cubic function between that and the x axis was 16. And if you subtract those 48-16

is an area of 32 which was almost the complete answer I believe…the area between those

two graphs because this is just a small area. Now this is a CAS version which stands for

Computer Algebra System. I could come over here to the, hit the tab button to toggle

between the two views, I could go to Menu- Calculus- Integral, my lower bound again was

-2 my upper bound was +2 and here I can just get the answer of 32. Hopefully get the same

answer. And actually I can type in the subtraction of those two function, now again the linear

function was first which was 4x +12 , minus, the other function has a lot of terms in it

so I am going to use parentheses . That cubic function was x^3 +3x^2. And we are going to

integrate that with respect to x. And voilà, we have 32 again. Now that was just one of

the two areas, you have to find the other one with a similar fashion and add those two

together to get the answer that I got in my video. Alright… thank you for watching.

Now let’s get back on to the next example. Third Example – Same Directions. Let’s

find the area bound by these two graphs. Again we have a straight line. So… in here we

have y^2=2x + 10. So we have clearly what is a parabola. And the y is squared instead

of x so it opens sideways. Sounds similar to something I talked about earlier in this

video. That means we have y^2=2(x+5). So our vertex here is at (-5, 0) since this is

already in standard form. SO (-5,0), we got that right there, if we let x=0, we are going

to have 2 times 5 is 10 and then y squared -10 so y is + or – square root 10 which is

a little more than 3. So our y intercepts are just a hair greater and smaller than 3

and our parabola opening sideways looks something like this, kind of. And then we have this

line of y=x+1. That clearly just has a y intercept of1 and a slope of 1. So this not really graph

paper, so I am just trying to estimate where all these little tick marks are. And so we

are looking for this area here. Well I have already discussed in this video, that why

I would not want to find or do a problem like this with vertical representative rectangles

because in this area here. I am going to do, well find the vertex but it’s only a

parabola. So not too big a deal there. We already did. But finding the area to the left

of that line is going to be complicated. So we are going to do this problem using horizontal

representative rectangles. So we are going to be basically using stuff like this sort

of. Right? With a dy. So we are going to get these functions rewritten so that x is in

terms of y. So here, well that’s not going to be too hard, we are going to write this

function as x=y-1 and we can call this function as g (y). g(y)=y-1. And then for our function

here, we can take this and subtract both sides by 10 and get 2x=y ^2 -10 and then divide

both side by 2 giving us x=½ (y^2-10). Lets go ahead and call that f(y). Okay, so now

I have my two functions in terms of y which will allow me to set up my definite integral

expression and be able to integrate with respect to y. But what are we going to use as our

lower and upper limits? Well we are going to take these functions and it really doesn’t

matter what format we look them in, but we do need to set this equation equal to 0 or

equal to themselves and solve those out and I don’t think we need too much help solving

systems of equations so I am going to step out and reveal that solution one step at a

time. Alright… you saw what I did, substitution, took that value of x from this 1st equation

plug it into the second one and solve for y and get y=4 and -2 which means our intersection

points, not that we really need to know the x’s because we are integrating with respect

to y, but they are (-3, -2) and (+3, 4). Well I am still finding out why the dogs are barking!

Alright now that my little watchdog here has made sure that we are safe let’s go ahead

and finish up setting the definite integral. Go on… guard the house!! Alright. So we

have, as we rise from our low y values to our high y values and I look at what’s bounding

our area here on the right side and the left side- that’s our greater x values and our

lower x values. Then this is going to be, area is equal to the definite integral from

lower limit of -2 to 4 and what are we subtracting again – the farther right function giving

us the greatest x values and that’s the straight line with independent variable of

y, of y-1 which I call g(y) minus our function on the left side that will be lower x values

in this bound area I am going to take this – ½ and distribute it through and get, how

about we call it, -½ y^2-5, and again find that integral with respect to y. And we should,

by now, have tons of practice of finding definite integrals. So I am going to go out and reveal

this one step at a time. If you want to try it on your own, just pause the video and work

out your answer before you see my mine. So our area came out to be 18 square units. If

you wanted to check this answer with a calculator, and I don’t know how to do numerical integration

on a calculator unless it’s with respect to x, if anyone knows any other method, may

be drop me a line and give me a hint on how to do that, but if I wanted to find this area

with a calculator I would have to get that parabola written so it would be y in terms

of x and you can see different colors here, you will have to find the area of this region

bound between the parabola and the x axis, the area of this region which is only the

triangle so that wouldn’t be too terribly hard how to figure out. That would be for

areas below the x axis and make sure that you wouldn’t have negative areas. Then you

would have to find the area of this region and of course we will have to find that value

here and they have lower and upper boundaries of -5 and -1 for that particular case and

then find the area of this one where you will have to find the area bound between parabola

and x axis between -1 and 3 and subtract it with the area bound between the straight line

and the x axis to find that final answer. So it would be quite a bit of work to find

the same answer of approximately 18 with the aid of calculator because you will have to

get it all in terms of y and in terms of x and find all these pieces individually. So

anyway that’s how I think of doing this problem. You got one more example dealing

with trigonometric function and odd functions. Hope you are finding this helpful. Last example,

same thing, find the area bound by the functions and you are only given two function and we

have a closed interval. So let’s see if these functions intersect… may be one function

is above the other one on the entire close interval… I don’t know. Don’t know until

we have a graph. So hopefully we remember graphing some basic trig functions. If not,

I will at least help a little bit here. We have sine function here whose only transformation

is an amplitude of 2 so instead of going like the sin (0)=0 and sin (pi/2)=1. Well sin(pi/2)

is 1 so 2 times the sin(pi/2) is +2. And we have a tangent function which has got a vertical

stretch of the square root of 2. So basically, we know, hopefully tan(0), that is y/x so

tan(0)=0/1 which is 0 and the tan(pi/4)=(square root of 2) /2 over the (square root

of 2) /2 which is 1. Well now with that amplitude change there, if you will, the vertical stretch

of value of square root of 2, would be square root of 2 instead of 1. So let’s just get

a quick sketch. Now the tangent function has asymptotes, there’s no horizontal transformation,

there’s no left or right phase shift, so it’s still going to be centered around the

origin and there’s also no vertical shift so, the standard period of tangent function

is pi. But the y axis, if you will, takes the first period and sort of splits it in

half, so we are going to go –pi/2 and roughly try to make this equal… +pi/2. And the tan(pi/2)=y/x.

x=0 so it is undefined. Thus we have our vertical asymptotes or at least two of the infinite

number of asymptotes. And then we know from my class last year that if you want to graph

trig functions, using or getting the simplest values possible, you count by one quarter

of the period and ¼ of the standard period of pi is pi/4. So tan(0)=0. tan(pi/4) is normally

1 but with the square root of 2 over here, that ‘A’ value of square root of 2. I

am just going to put that here and that means the function is going to look something like

this…make the point bigger so I actually hit it…and that tangent function is going

to come down through to pi/4- square root of 2, come through (0,0) and of course hopefully

you remember some properties of trig functions- that tangent function, barring the fact that

it has shifted left or right or up or down of that origin, is an odd function, so it’s

going to look something like that. Probably made the space a little bit too big! But,

it kind of swings down and sort of makes an S shape through the origin. Okay. Now the

sine function, at least within the area that we have there. sin(0)=0 and 0 times 2 is 0.

sin(pi/4) is a little bit of a nasty problem, but I do know that sin(pi/2) is equal to 1

and now it’s going to be multiplied by 2. So the sine function is going to look something

like this. And the sin(-pi/2) is -1, but no, now it’s -2. Very rough sketch, I am drawing

here. I apologize for my sloppiness. We are going to need to find the area of these two

regions. Now, are those really intersecting at pi/4 and –pi/4 or does it fall a little

bit shy or is the intersection point a little bit to the right of pi/4 and so the orientation

of the graph flip flops? Because here the sine is above tangent function, if the intersection

point is before pi/4, then they are going to switch orientation and we are going to

have to do another integral with the subtraction reversed so that we don’t get a negative

area. Well, I have a drawing done, you could see it, a very rough sketch. I am kind of

making it look like that the intersection points are happening at x values of pi/4 and

–pi/4 but we don’t really know that. All we know is that on this side of quadrant 1,

the sine is above the tangent function and in quadrant 4, the tangent function is above

the sine function. So let’s work on finding the intersection points. Well that means we

are going to have to set these equal to zero…or set them equal to themselves. So it will become,

set equal to zero. We have an equation with 2 variables- we can’t really solve an equation

usually when there’s two variables in it. So really, we are not going to be able to

do anything here, but with us having a sine and tangent function. SO I am going to do

two things, I am going to move this term over here to the left and I am also going to write

it in terms of sine and cosine. Okay, now, when you are talking about algebraic functions,

maybe there’s something more complicated that I don’t have over the top of my head

or we haven’t really dealt at the level of math we are at. One of the rare cases when

we can solve an equation when there’s two variables in it at the same time, is if they

factor apart. We see both of these terms have a factor of sine and there’s a cosine function

too. So still we can have two variables in the same equation but the sine, if you take

that out, we get sin(x) times [2- (square root of 2)/ cos(x)]=0. Now we have factored,

now we have our two variables in two separate factors which is good because after you factor

an equation, of course, you set each one equation to 0 and voila, our two variables are now

in two separate equations. This video is a little bit long and I really don’t know

if you’re willing to see me talk through every step of the solution of these individual

equations for x. I am going to step out and reveal those answers and get this graph out

of the way. Alright, solving each one of those little equations, we had x=0 and we have an

x values of –pi/4, +pi/4. So the author here, or I wasn’t trying to trick you. This

is my problem that’s inspired by a textbook problem. So without trying to trick you too

much, we have an intersection point as (0, 0) and the graphs intersect at the lower and

upper limits of our closed interval. So, before we finish this up, now I get to the main end

of the problem where I need to find the area. This again, the tangent function, assuming

you haven’t moved it left-right or up or down so that you ruin that symmetry that it

has, so in the sort of parent form, you just have a change in the amplitude, if you will,

tangent doesn’t really have an amplitude. Amplitude goes up and down forever. But this

is an odd function. If you plug in 2 opposite numbers like –pi/2 and +pi/2, you get opposite

answers. This is an odd function. Well, the sine function, barring any issues with phase

shifts or vertical shifts, this is just still sort of centered around the origin except

for the amplitude change. It’s an odd function as well so since both of these are odd functions,

we can find just one of these areas and multiply it by 2. So as I get all this out and cleaned

up, I am going to set up a definite integral from 0 to pi/4 but I am just going to do that

from 0 to pi/4 and multiply it by 2 so that we don’t have to find the area of this region

where the tangent function is above the sine function and then turn around and find the

area of this region where the sine function is above the tangent function. We can just

find one and multiply it by 2. So recognizing these times when you have even and odd functions

can save you some work on your problem. Just use all of those things you learnt on Algebra

2 and Pre- Calc. Alright! We got our definite integral set up. It is from the area in quadrant

1 multiplied by 2, again, to account for the one in Quadrant 3 that will be equal in area.

I am going to write this separately and have this as 2 times the definite integral from

0 to pi/4 times 2 sin x minus 2 times the definite integral from 0 to pi/4 of square

root of 2 tan(x) dx. And I just realized that I forgot my dx in there, don’t do that…your

teacher will probably mark off a couple of points. Now I am going to, in the next line,

take the 2 out front, we know the integral of sin(x) is –cos(x), dx slipped right in.

And I am going to bring this square root of 2 out front. And now if you remember the integral

of tangent function, great! If you don’t remember the integral of tan(x), I going to

write this as sin(x) /cos(x) just to walk you through an extra step if you need to,

because you forgot the integral of tan(x). But you really need to start memorizing those

integration rules for standardized test of for your tests or maybe for the AP exams coming

up in probably a few months if you are watching this video right now. SO this is going to

be 4 times the definite integral from 0 to pi/4of sin(x), let’s get that dx in there

now, minus 2 times the square root of 2 definite integral from 0 to pi/4… Now if we write

this tan(x) and sin(x)/cos(x) dx, we can recognize, hopefully an integration rule that will help

us finish this problem up. See if we let u be our denominator, remember that integrating

pattern of u prime over u. If we let u=cos(x), derivative of cos(x)=-sin(x). We have a

positive sign up front so if we introduce a negative and then a negative 1 out front…essentially

multiply and divide by -1, I am going to change the look of this integral but not its value.

Now we have the pattern of u prime over u. The integral of u prime over u is some sort

of natural log function. So at this point, hopefully, you are starting to see how to

finish this problem up. So I am going to step out to speed things up a little bit and show

the solution step by step. BAM! Well maybe you might have done all this work and got

to this answer and go to the book or the multiple choices options you have on a test and realize

it doesn’t match. Okay…if you were doing your homework, did you do it right? Well you

can plug it into the calculator and see if you get the same decimal approximation as

the book. If you are taking a test and don’t have access to a calculator, that’s not

going to help, or again, the multiple choice test, not getting that in the format of one

of the choices so this is done and unless I have made a mistake, it’s all correct.

But how can we change the look of this to again, possibly match the answer at the back

of the book or maybe actually with what came out of my TI-INSPIRE CAS…with a Computer

Algebra System or with the choices on multiple choice test. We have all these properties

that we can use to expand log functions and expressions and I can rewrite this as -2 times

the square root of 2 +4 + 2 times the square root of 2 times the natural log of 2 to the

one half power over 2 to the 1st. 2 to the ½ power is of course square root of 2. When

you divide like bases, you subtract the exponents. SO 2 to the one half power over 2 is 2 to

the – ½ power. I am rationalizing the denominator here. Alright! Well..when you take the log

of something that has an exponent, the power property allows you to take that exponent

and pull it out front. When you do that, you’re going to have -½ times +2. Half of 2 is 1

and bring that negative and you have a final answer of -2 times the square root of 2 +4

minus, from that negative times positive and that 2 divided by 2 is 1, so you get minus

square root of 2 times the natural log of 2. Now that might be the answer that you see

on a multiple choice test or in the back of your book. And here it was right as well.

If it’s an AP exam, it’s an open ended response, you will get points for simplifying

your answer. Maybe that’s acceptable, I don’t know…I hope I have shown you enough

to get you through your homework. So I am Mr. Tarrou. BAMM! Go do that homework!

Here is my lesson on calculating the area between 2 curves:)

this channel is the key to pass my exams

Thank you

An hour, wow, bravo! You don't see too many of us trying the truly cray cray on the internets. You ready for a challenge> U.S. History Final Exam Review: FOUR HOURS!!!!!!! #neveragain

Thank you for the in depth lesson! Your calculus videos help a lot and supplement learning in the classroom! Keep up the good work!

Learn a lot and without struggling. To. Understand

hello sir. please help. how do I integrate y=x^3, x=0,x=2 with respect to y? the x's are confusing me. please help. thanks.

I am wanting your arc length formula video for calc 2. as well as the others. will you be posting vector calc vids as well or am I just dreaming. you are great. you should teach physics as well. you have a great teaching style. have a great night and….BAMMM thank you very much. keep those vids coming.

Enlightening! 🙂 A thousand splendid thanks!

You can do integration on your calculator when you have an integral related to y. Just enter the same bounds and equation but use x instead of y as your variable and you will get the correct answer.

Hi sir please can u make another channel for people who are in secondary schools by the way I live in england

Thanks for the help again Prof! I think on your last example there was a tiny mistake at the very end, where 4 x -1 was written as +4, instead of -4. It's in the last 2, 3 mins of the video. Again, thanks for the video! It's a life saver for me!

I get a lot of help from your lessons. They are very clear and easy to learn from. Thank you.

23:22

x minus 6*

Walked out of class today with my brain almost suffocating in confused integrals/trig functions. So I came home, watched this video, and now I can't thank you enough. This really cleared things up thanks prof!

It really helped me a lot 🙂 I have watched lots of video about this topic and still I don't understand then I found this one! Thank you Prof.Rob 🙂 I wish you teach here in Philippines. 🙂

your videos has to be some of the best videos on calculus. Thanks you so much for your help

@ProfRobBob Mr. Tarrou, you are da man! Just started my journey into Calc 2 in a summer course and plan on watching your videos every step of the way. Don't know what I would do without you…..BAM! HAVE A GREAT SUMMER!

Closed Captioning brought to you by my YouTube student and new friend from India, Jigyasa. Thank you again so much for the help:) #math

24.08

You forgot BAMMMMMM .. we were waiting for it

Test coming now, Syllabus is Application of integrals and Differential equations And Vectors .. Gotta prepare now .. tell you the result later

57:53 bahahhaha love the sound effects!

can you explain why, at 45:28 you set up the definite integral the way you did? If you look at the graph sideways, it appears the parabola is the greater function, with respect to the bounded interval?

a math prof??? with good handwriting???? what ?? ? ??

please solve this equation i trying to solve by myself but didn't get the answer please help me prof.

"sketch

and find the area of the region bounded by the curves y=sinx and y=pix-x2"

nice dog

I'm from libya and you're my teacher in calculus 1 and 2 thank you <3

IGot 40 in first exam calculus 2

+ProfRobBob I keep getting 125/4 instead of 131/4 in the problem at 34:10. I worked it out many times and I still get that answer. Help please!

This is better than my college Calc II class!! You got me through Pre-Cal/Trig, and Calculus I, with A's in both! Got a B on this first exam in Calculus II, but it was my fault….Just couldn't brain one whole problem until after I'd left the building. Can't say you didn't cover it though!

Thank you Mr. Tarrou! These lectures are killer!!

Hello professor is this chp 7.1 area of a region between two curves?

Hey Prof!

I spotted a really small mistake at 18:20; you just missed a pair of parentheses around g(y) when writing it under the integration thingy.

Fantastic video nonetheless!

it's absolutely superb

Thank you so much!! My prof slightly brushed on this topic, but you explained everything really well. Specifically, I was looking for an explanation on solving horizontally! 🙂

Aswom

At 46:31 when you were calculating the last step, for some reason you put 4(-1/2) instead of 4(-2) like you should've for that step x.x but you still put the correct answer overall of 18.

Also, thank you for your videos ProfRobBob. I'm going to be attempting calc 3 again in a week, and I'm watching all your calc videos to refresh myself on Calc 2.

Love your dog! Beautiful handwriting! As always, solid content! I recommend your video to my students! Thanks for your work!

He looks like Mr. Bean with good penmanship and mathematical skills eme