Find the Area of a Region Between Two Curves Using Integration

Video Series: Applications of Integration (Complete In-Depth Tutorials For Calculus)

(D4M) — Here is the video transcript for this video.

00:00
in this video you’ll learn how to find the area of a region between two curves
using integration did you know that finding the area
between two regions is basically the same as finding the area separately and
then subtracting the areas thus this topic makes a good review of the
fundamental theorem of calculus so let’s get started
hi everyone welcome back in today’s video find the area of a region between
two curves using integration is part of the series applications of integration
complete in-depth tutorials for calculus
let’s find out what we’re going to cover in today’s video
so first we’re going to talk about the area between two curves
and then we’re gonna do example one example two and so we’re gonna do three
examples to make sure that we understand
how this process works and also to get a really good review of the fundamental

00:01
theorem of calculus we’re going to be using that in all of our examples
so let’s get started so uh first is what exactly do i mean by
area between two curves and so let’s do um
an illustration here to understand here what’s going on
so what we’re gonna have is two continuous functions
and we’re gonna have a closed bounded interval
and we’re gonna say that f dominates g so throughout the interval here
f is greater than g so we can sketch the graph over here of some
uh example or some illustration here so let’s look at say you know we have this
function nice and continuous let’s say this is a graph of f
and let’s say this is the graph of g right here now i notice i drew
g above f over here but here’s where our a and b are going to be a and b

00:02
so we have two continuous functions and if we only look at this interval here
between a and b here so on this interval right here g is below the f
and so here we have this vertical lines here uh x equals a and x equals b
and so we’re asking here what is this area in here
we can find this area right here and we can do this
two different ways we can do it this way right here or
if we want we can expand this out and say this is the integral from a to b of
f of x my dx minus the integral from a to b of g of
x so if we see what this right here is this is the area right here under f

00:03
which is all of this the red and the below part right here and the area under g
the graph of g is just this area down here and but we’re subtracting it away
so we’re left with only the area here in red
so this is how we find the area between two curves is just find the area
separately and then subtract of course you could subtract the functions first
and then integrate so that’s usually the way we do this if we can is we’ll
subtract the functions first and get one function here and then go ahead and
integrate so let’s look at some concrete examples to to get uh this going here
so example number one here we go so find the area enclosed um by this
parabola and this line right here so y equals minus x
2 minus x squared and so let’s get a let’s get a sketch going here so this is
the parabola x square but it’s been turned upside down and it’s been shifted

00:04
up to so that’s that one’s pretty easy to graph and then we have the line
slicing right through there negative x so let’s see if we can get a good sketch
going um let’s say we have here the axes right here
and we’re going to have this parabola here
and let’s say that’s good enough there and then we’re going to have this line
chopping right through here y equals minus x so this is y equals minus x
and this is the parabola right here two minus x squared
and we’re looking for this uh area in here the area between these two curves
alright so there we go so what we need is to to integrate this we need to know
where this intersection point is here and where this intersection point is
here so we need to find these two x intersection points because this one
right here will give me my a and this one right here will give me my

00:05
b and we’re going to integrating from a to b here
so to find these two intersection points well they’re both equal to y
so it should be straightforward so this is y
and that’s equal to y that’s also equal to y minus x
so i’ll just move the x squared over and i’ll move the minus 2 over i’ll put
everything on the right hand side and then i’ll flip sides so i’ll say this is
x squared minus x and then a minus two all right very good and then we can
factor this x minus two x plus one [Music]
and so we’re going to get x is 2 and x is minus 1.
and now for this x here what do we get for the y when x is 2 y will be minus 2
and then when x is minus one the y will be positive one

00:06
so here are two points here so this will be the point right here
let’s go ahead and label this right here this will be the point when the x is
positive two and the y is minus two and this will be the point right here
when the x is negative negative one one so that point right there is negative
one one all right so we set up um all the work we needed
to do to to now we can integrate to finding to find the area
so we can do the area is the integral from
um the x the a is minus one and here the x is two so minus one to two
and we’re going to do the um now but while you’re setting this up it’s a good
idea to double check that we have an a continuous upper curve the function here
is f and that’s the upper curve for the whole

00:07
region so in other words when i start drawing when i start drawing strips no
matter where i draw a strip the upper curve is this f right here so
this is my f of x and the lower curve here all the way
through the whole interval minus one to two this is my lower curve right here
and this is my g of x here so we’re going to have f of x which is two minus x
squared minus our g of x here which is a minus x so all of that right there dx
so this will be our f of x minus the g of x here
and this right here will give us the area in red here
now to find this integral right here we can combine the f and g first
so let’s say this is a minus x squared and this is plus x

00:08
and then this is plus 2. okay so now it becomes just a regular
fundamental theorem of calculus problem or in other words just finding the
definite integral here so i’m looking at minus x squared so i’m going gonna do
minus x to the third over three plus x squared over two plus two x
and then we’re going to evaluate minus one to two okay good so
now let’s go up here to get this worked out here let’s go down here let’s say
and substitute in a two everywhere so this will be minus eight thirds plus
and then that’s four over two so that’s two and then plus a four
now let’s substitute in minus one everywhere so this will be a minus one
to the third so that’ll be uh what negative negative so that’ll be one-third

00:09
and then this will be plus one-half and then that’ll be minus two
and then now we’re going to get all of these fractions here we’re gonna get a
minus positive put the minus in front so we’re
going to subtract all that out and we get in fact nine over two
just putting those fractions together and simplifying so that’s the area
in the red here is nine over two and then if you want to put units you
can just say nine over two units squared unit squared
although most people i mean i don’t know most people but i’m just going to leave
it as 9 over 2. all right so there’s our first example there so i
thought it was fun to find the intersection points just sketch a graph
so that we can kind of see what’s going on but in the end the heavy work is
integrating and that’s just the fundamental theorem

00:10
of algebra uh calculus right there all right so let’s look at another example
example two and here we go so find the area of the shaded region
and what do we have here we have this graph here x to the fourth minus 2x
squared that’s got some symmetry in it because they’re both even powers
and then we have a minus uh so we have a 2x square so that that
one’s pretty easy to graph and then this one right here is fourth power here so
in order to sketch the graph really good either you have a good graphing
calculator or maybe you can do some um calculus one
here where you find the minimums and the
maximum where it’s increasing where it’s
decreasing so however you want to uh you know technique you want to use to graph
that but in the end you need a you know decent graph so that we can understand
what’s going on so i’m going to um sketch the graph over

00:11
here and i’m going to try to sketch this one right here first to get the good
shape of it so it’s going to come down something like this it’s going to come
down like this and then go back up and it’s going to come down and it’s
symmetrical and then it’s going to go up it’s going to look something like that
and then this one right here is just going to come through here like this
so it’s going to come like that and these should be about the same height here
let’s go up a little bit higher it should come up about right here
all right so there’s the graph of so this is this one’s the
um so which one’s my upper curve throughout this region here so we’re
looking at defining this area here let’s identify the area here

00:12
we’re looking this is the area here we’re trying to find in red
and what is the upper curve um so this will be the upper curve right
here so i’m going to say f of x equals the 2x squared
and this is the lower curve here the g of x which will be the x to the fourth
minus 2x squared this is my x-axis my y-axis
all right so so far so good yeah so we need to find these intersection points
here and we don’t really need to know where the minimums are um
because we’re going to be integrating here and in fact because this problem is
symmetric i’m going to use a symmetry here and actually i’m only all i’m
really going to do is find the area of the one on the right and then double
that area to get the area of the whole thing so i want to be integrating from 0
to and then we need to find out what is this intersection point here
so let’s see if we can do that over here so again they’re both equal to y

00:13
so on one hand we have uh y is x to the fourth minus two x squared
and on the other hand the y is 2x squared
so if i move this over here i have x to the fourth minus 4x squared equals zero
and we can factor out an x squared so x squared minus four
and you can see that the zero here it bounces at zero it’s an even power and
then we have plus or minus two that works right here so x equals zero
x equals minus two and x equals two those are the
intersection points there but we’re really interested in this x equals two
right here so this is a two and that’s the point right there is two
and then what’s the height here we could go find the height although it’s not
really necessary to find the area the height will be two squared times two
which is an eight okay so we just really need the two

00:14
we’re going to integrate from zero to two now let’s double check here before
we set this integral up that all the way through the interval here
we have as we’re taking strips here that this is the upper curve all the way
throughout the interval zero to two and then this is the lower curve g of x
here all throughout the interval here and so since that’s happening we can use
the area to find the formula so the area is um so i’m going to use a um 2
to double the area and then i’m going to integrate from 0 to 2
and i’m going to have my f of x which is 2x squared
minus my g of x and i better put parentheses around this g of x here to
get the whole g of x so f of x minus g of x dx so that’s the area right there

00:15
once we integrate this out we get the area as a number
so right now we can just simplify f minus g of x
and say this is zero to two and this is going to be minus x to the fourth and
this will be positive x 2x squared so this will be another 2x
squared here so i’m going to get a positive 4x squared here dx
all right very good so now we can integrate this right here and
we’re going to get here x to the minus x to the fifth over five
and then plus four x to the third over three okay integrating each term
and we still have a two out front i’m going to evaluate this from zero to two
all right so very good so now we need to substitute into everywhere

00:16
and the the really useful thing about the symmetry in this problem is that
this zero here when you substitute it in
it all vanishes right here so we’re what we’re going to end up with is 2
and then we’re going to end up with a minus 2 to the fifth so minus 32 over 5
and then 2 to the third is 8 times 4 is 32 so 32 over 3
and then again when we substitute it into zero is we have an x here and an x
here it’s all going to um just add up to zero
so we just need to find these fractions right here and in fact when you put it
all together you’re going to get 96 over five
and so that’s the area in the red the total area right there we already
doubled it so that’ll be the total area in red right there so that’s a five
all right so very good there’s our second example right there
and let’s go on and look at another example now

00:17
all right so example three here we go so example three here we’re going to be
looking at this cubic here and a quadratic so let’s get this
cubic here and the quadratic here and i have an extra y equals in there
so just disregard that right there so here’s our y equals here our y equals here
and i need to sketch the graphs we have this cubic
so again to graphic cubic you can either use a calculator
um you can use calculus one uh because this is really a calculus one
at the end of the course topic or calculus two topic see the way you
should know first derivative test or second derivative test or something like
that you can figure out where it’s increasing decreasing concavity and all
that stuff but in the end you want a nice good shape
and so i’m going to get my cubic going so it’s going to come up like this

00:18
and then it’s going to come down to the origin here
and then it’s going to come back up through here like this
that’s going to be my cubic so as x goes to positive infinity for
example it’s just going to blow up and as x goes to
negative infinity it’s just going to keep falling down right here
all right so here’s my x-axis here’s my y-axis and then i have my
upside down parabola but it’s been shifted a little bit so it’s not going to go
it actually still goes to the origin zero zero so it’s going to just kind of
come up like that so it’s gonna say come up like
get a good sketch here something like that
and so this is gonna have to come back down and we’re looking for these two
intersection points here this one is definitely not symmetric you can see
that right here this is not symmetric with respect to the y-axis so it’s just
some parabola that’s coming up and then coming down somewhere
and we need these two intersection points here

00:19
now when i’m setting this one up what i notice is that the bounded region right
here is really this is the area that we’re after right here
as you see in these two regions right here
these two parts right here when i when i start taking strips right here
i have a different lower and a different upper so when i’m taking strips over
here on this part this cubic here is the upper and the parabola is the lower
now when i start taking strips over here to when i start integrating over here
now the quadratic is the upper and the cubic is the lower so we’re gonna have
to break this area up here in red into two smaller areas i’m gonna find the
area of this one in red here on the left and i’m gonna find the area of this one
over here on the right so let’s do that so in order to do that of course we need

00:20
to find the intersection points here so let’s do that over here first
so on one hand we have the y which is 2x to the third
minus x square minus five x and then on the other hand y is minus x
squared plus three x okay so now let’s put everything on one
side so this will be 2x to the third and let’s move this minus x
squared over we’re going to get positive x squared so the x squares are going to
add up to 0. and let’s move the 3x over here and we’re going to get minus 8x
and we can factor out a 2x and we’ll get an x squared minus a 4
right so 2x to the third and then minus 8x good so we get the x equals 0 which
is right there and then we have a difference of two
squares again so we’re gonna get minus two and two

00:21
and so this one right here is a two and this one right here is a minus two
okay so we need to integrate from minus two to zero to find this area in red and
here we need to integrate from zero to two to find this area in red here and so
we’re gonna set that up here as two areas so the area is
two integrals the first one will go from minus two to zero integrating here from
minus two to zero now here the upper curve is the cubic so i’m going to write
the cubic first so bracket and then parentheses so we have 2x to the third
minus x squared minus 5x minus the lower curve in this problem on
this side right here the lower is the quadratic
which is minus x squared so i need parenthesis minus x squared plus 3x

00:22
close bracket and then dx so this integral right here is just for the area
on the left so plus now the area on the right
so now we’re going to integrating from zero to two
bracket and we’re going to subtract functions now the upper curve here is
the quadratic so this one will come first so minus x squared plus 5x
minus and now the lower curve which is the cubic for this one right here
which is the two x to the third minus x squared minus five x and then dx
all right so pr oh i forgot my bracket close my bracket off here and then dx
all right so you see how they’re different over here um
and that’s why i think it’s a really good idea to sketch a graph and have it
have a really good decent graph here all right so now we want to integrate

00:23
each one of these now first thing we’ll do is we’ll subtract these together
and then we’ll integrate term by term and so we saw how to do that on the
previous example so in this example right here i’m just gonna say for this
first integral right here you’re gonna get an eight and then plus
and then for the second integral right here integrating all this out right here
you’re also gonna get an eight and so in the end you’re gonna get a sixteen
so that’ll be the answer so these integrals both come out to eight
so even though they’re uh don’t look like the same shape
they actually have the same area and so they’re both eights and then so
the total area all of it in red is the 16 right there
so there we go i hope you enjoyed that example as well
and so double check that you get that 16. now if you have any questions or
ideas please uh leave a comment below if

00:24
you like this video go ahead and like it don’t forget this video is part of the
series applications of integration complete in-depth tutorials for calculus
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David A. Smith (Dave)

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David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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