# Area Between 2 Curves Vertical and Horizontal Representative Rectangles Calculus 1 AB 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!

### 37 thoughts on “Area Between 2 Curves Vertical and Horizontal Representative Rectangles Calculus 1 AB”

• January 19, 2014 at 1:28 pm

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

• January 21, 2014 at 10:35 pm

this channel is the key to pass my exams
Thank you

• January 21, 2014 at 11:34 pm

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

• January 23, 2014 at 3:48 am

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

• January 27, 2014 at 3:18 pm

Learn a lot and without struggling. To. Understand

• February 4, 2014 at 2:45 am

• February 11, 2014 at 6:20 am

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.

• March 4, 2014 at 1:30 am

Enlightening! 🙂 A thousand splendid thanks!

• May 4, 2014 at 5:16 pm

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.

• September 25, 2014 at 6:06 pm

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

• November 25, 2014 at 3:42 pm

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!

• January 29, 2015 at 4:49 am

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

• February 2, 2015 at 2:58 am

23:22
x minus 6*

• February 5, 2015 at 12:42 am

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!

• February 17, 2015 at 9:44 pm

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. 🙂

• May 14, 2015 at 1:11 pm

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

• May 29, 2015 at 2:23 am

@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!

• May 31, 2015 at 6:25 pm

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

• August 7, 2015 at 7:32 pm

24.08

• October 22, 2015 at 2:24 pm

You forgot BAMMMMMM .. we were waiting for it

• October 22, 2015 at 2:26 pm

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

• December 17, 2015 at 12:39 am

57:53 bahahhaha love the sound effects!

• January 29, 2016 at 10:05 pm

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?

• April 18, 2016 at 9:57 pm

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

• April 30, 2016 at 9:16 am

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

• September 2, 2016 at 3:10 am

nice dog

• December 20, 2016 at 9:30 am

I'm from libya and you're my teacher in calculus 1 and 2 thank you <3
IGot 40 in first exam calculus 2

• February 4, 2017 at 12:32 am

+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!

• March 8, 2017 at 3:16 am

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!!

• March 8, 2017 at 7:47 pm

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

• April 27, 2017 at 7:38 pm

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!

• April 29, 2017 at 6:49 am

it's absolutely superb

• July 4, 2017 at 11:58 am

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! 🙂

• July 23, 2017 at 5:57 pm

Aswom

• August 16, 2017 at 1:36 pm

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.

• February 6, 2019 at 5:52 pm
• 