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If R is the region bounded by the graphs of the functions f ( x ) = x 2 + 5 and g ( x ) = x + 1 2 over the interval [ 1 , 5 ] , find the area of region R .

12 units 2

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In [link] , we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.

Finding the area of a region between two curves 2

If R is the region bounded above by the graph of the function f ( x ) = 9 ( x / 2 ) 2 and below by the graph of the function g ( x ) = 6 x , find the area of region R .

The region is depicted in the following figure.

This figure is has two graphs in the first quadrant. They are the functions f(x) = 9-(x/2)^2 and g(x)= 6-x. In between these graphs, an upside down parabola and a line, is a shaded region, bounded above by f(x) and below by g(x).
This graph shows the region below the graph of f ( x ) and above the graph of g ( x ) .

We first need to compute where the graphs of the functions intersect. Setting f ( x ) = g ( x ) , we get

f ( x ) = g ( x ) 9 ( x 2 ) 2 = 6 x 9 x 2 4 = 6 x 36 x 2 = 24 4 x x 2 4 x 12 = 0 ( x 6 ) ( x + 2 ) = 0.

The graphs of the functions intersect when x = 6 or x = −2 , so we want to integrate from −2 to 6 . Since f ( x ) g ( x ) for −2 x 6 , we obtain

A = a b [ f ( x ) g ( x ) ] d x = −2 6 [ 9 ( x 2 ) 2 ( 6 x ) ] d x = −2 6 [ 3 x 2 4 + x ] d x = [ 3 x x 3 12 + x 2 2 ] | −2 6 = 64 3 .

The area of the region is 64 / 3 units 2 .

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If R is the region bounded above by the graph of the function f ( x ) = x and below by the graph of the function g ( x ) = x 4 , find the area of region R .

3 10 unit 2

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Areas of compound regions

So far, we have required f ( x ) g ( x ) over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.

Finding the area of a region between curves that cross

Let f ( x ) and g ( x ) be continuous functions over an interval [ a , b ] . Let R denote the region between the graphs of f ( x ) and g ( x ) , and be bounded on the left and right by the lines x = a and x = b , respectively. Then, the area of R is given by

A = a b | f ( x ) g ( x ) | d x .

In practice, applying this theorem requires us to break up the interval [ a , b ] and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.

Finding the area of a region bounded by functions that cross

If R is the region between the graphs of the functions f ( x ) = sin x and g ( x ) = cos x over the interval [ 0 , π ] , find the area of region R .

The region is depicted in the following figure.

This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled “R1” and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled “R2” and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.
The region between two curves can be broken into two sub-regions.

The graphs of the functions intersect at x = π / 4 . For x [ 0 , π / 4 ] , cos x sin x , so

| f ( x ) g ( x ) | = | sin x cos x | = cos x sin x .

On the other hand, for x [ π / 4 , π ] , sin x cos x , so

| f ( x ) g ( x ) | = | sin x cos x | = sin x cos x .

Then

A = a b | f ( x ) g ( x ) | d x = 0 π | sin x cos x | d x = 0 π / 4 ( cos x sin x ) d x + π / 4 π ( sin x cos x ) d x = [ sin x + cos x ] | 0 π / 4 + [ cos x sin x ] | π / 4 π = ( 2 1 ) + ( 1 + 2 ) = 2 2 .

The area of the region is 2 2 units 2 .

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If R is the region between the graphs of the functions f ( x ) = sin x and g ( x ) = cos x over the interval [ π / 2 , 2 π ] , find the area of region R .

2 + 2 2 units 2

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Finding the area of a complex region

Consider the region depicted in [link] . Find the area of R .

This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The region is labeled R. The shaded area is between x=0 and x=2.
Two integrals are required to calculate the area of this region.

As with [link] , we need to divide the interval into two pieces. The graphs of the functions intersect at x = 1 (set f ( x ) = g ( x ) and solve for x ), so we evaluate two separate integrals: one over the interval [ 0 , 1 ] and one over the interval [ 1 , 2 ] .

Over the interval [ 0 , 1 ] , the region is bounded above by f ( x ) = x 2 and below by the x -axis, so we have

A 1 = 0 1 x 2 d x = x 3 3 | 0 1 = 1 3 .

Over the interval [ 1 , 2 ] , the region is bounded above by g ( x ) = 2 x and below by the x -axis, so we have

A 2 = 1 2 ( 2 x ) d x = [ 2 x x 2 2 ] | 1 2 = 1 2 .

Adding these areas together, we obtain

A = A 1 + A 2 = 1 3 + 1 2 = 5 6 .

The area of the region is 5 / 6 units 2 .

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Questions & Answers

why n does not equal -1
K.kupar Reply
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
Darnell Reply
proof the formula integration of udv=uv-integration of vdu.?
Bg Reply
Find derivative (2x^3+6xy-4y^2)^2
Rasheed Reply
no x=2 is not a function, as there is nothing that's changing.
Vivek Reply
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
Joys Reply
y=800
Gift
800
Bg
how do u factor the numerator?
Drew Reply
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Kranthi Reply
answer please?
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Lerato Reply
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
Amna Reply
what is a function? f(x)
Jeremy Reply
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
jon Reply
is x=2 a function?
The
What is limit
MaHeSh Reply
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
Andrew Reply
what is implicit of y³+x²y³=5 at (2,1)
Estelita Reply
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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