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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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