# 6.1 Areas between curves  (Page 2/4)

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

$12$ units 2

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 $f\left(x\right)=9-{\left(x\text{/}2\right)}^{2}$ and below by the graph of the function $g\left(x\right)=6-x,$ find the area of region $R.$

The region is depicted in the following figure.

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

$\begin{array}{ccc}\hfill f\left(x\right)& =\hfill & g\left(x\right)\hfill \\ \\ \hfill 9-{\left(\frac{x}{2}\right)}^{2}& =\hfill & 6-x\hfill \\ \hfill 9-\frac{{x}^{2}}{4}& =\hfill & 6-x\hfill \\ \hfill 36-{x}^{2}& =\hfill & 24-4x\hfill \\ \hfill {x}^{2}-4x-12& =\hfill & 0\hfill \\ \hfill \left(x-6\right)\left(x+2\right)& =\hfill & 0.\hfill \end{array}$

The graphs of the functions intersect when $x=6$ or $x=-2,$ so we want to integrate from $-2$ to $6.$ Since $f\left(x\right)\ge g\left(x\right)$ for $-2\le x\le 6,$ we obtain

$\begin{array}{cc}\hfill A& ={\int }_{a}^{b}\left[f\left(x\right)-g\left(x\right)\right]dx\hfill \\ & ={\int }_{-2}^{6}\left[9-{\left(\frac{x}{2}\right)}^{2}-\left(6-x\right)\right]dx={\int }_{-2}^{6}\left[3-\frac{{x}^{2}}{4}+x\right]dx\hfill \\ & ={\left[3x-\frac{{x}^{3}}{12}+\frac{{x}^{2}}{2}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{6}=\frac{64}{3}.\hfill \end{array}$

The area of the region is $64\text{/}3$ units 2 .

If R is the region bounded above by the graph of the function $f\left(x\right)=x$ and below by the graph of the function $g\left(x\right)={x}^{4},$ find the area of region $R.$

$\frac{3}{10}$ unit 2

## Areas of compound regions

So far, we have required $f\left(x\right)\ge g\left(x\right)$ 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 $f\left(x\right)$ and $g\left(x\right)$ be continuous functions over an interval $\left[a,b\right].$ Let $R$ denote the region between the graphs of $f\left(x\right)$ and $g\left(x\right),$ 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

$A={\int }_{a}^{b}|f\left(x\right)-g\left(x\right)|dx.$

In practice, applying this theorem requires us to break up the interval $\left[a,b\right]$ 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 $f\left(x\right)=\text{sin}\phantom{\rule{0.2em}{0ex}}x$ and $g\left(x\right)=\text{cos}\phantom{\rule{0.2em}{0ex}}x$ over the interval $\left[0,\pi \right],$ find the area of region $R.$

The region is depicted in the following figure.

The graphs of the functions intersect at $x=\pi \text{/}4.$ For $x\in \left[0,\pi \text{/}4\right],$ $\text{cos}\phantom{\rule{0.2em}{0ex}}x\ge \text{sin}\phantom{\rule{0.2em}{0ex}}x,$ so

$|f\left(x\right)-g\left(x\right)|=|\text{sin}\phantom{\rule{0.2em}{0ex}}x-\text{cos}\phantom{\rule{0.2em}{0ex}}x|=\text{cos}\phantom{\rule{0.2em}{0ex}}x-\text{sin}\phantom{\rule{0.2em}{0ex}}x.$

On the other hand, for $x\in \left[\pi \text{/}4,\pi \right],$ $\text{sin}\phantom{\rule{0.2em}{0ex}}x\ge \text{cos}\phantom{\rule{0.2em}{0ex}}x,$ so

$|f\left(x\right)-g\left(x\right)|=|\text{sin}\phantom{\rule{0.2em}{0ex}}x-\text{cos}\phantom{\rule{0.2em}{0ex}}x|=\text{sin}\phantom{\rule{0.2em}{0ex}}x-\text{cos}\phantom{\rule{0.2em}{0ex}}x.$

Then

$\begin{array}{cc}\hfill A& ={\int }_{a}^{b}|f\left(x\right)-g\left(x\right)|dx\hfill \\ & ={\int }_{0}^{\pi }|\text{sin}\phantom{\rule{0.2em}{0ex}}x-\text{cos}\phantom{\rule{0.2em}{0ex}}x|dx={\int }_{0}^{\pi \text{/}4}\left(\text{cos}\phantom{\rule{0.2em}{0ex}}x-\text{sin}\phantom{\rule{0.2em}{0ex}}x\right)dx+{\int }_{\pi \text{/}4}^{\pi }\left(\text{sin}\phantom{\rule{0.2em}{0ex}}x-\text{cos}\phantom{\rule{0.2em}{0ex}}x\right)dx\hfill \\ & ={\left[\text{sin}\phantom{\rule{0.2em}{0ex}}x+\text{cos}\phantom{\rule{0.2em}{0ex}}x\right]\phantom{\rule{0.2em}{0ex}}|}_{0}^{\pi \text{/}4}+{\left[\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}x-\text{sin}\phantom{\rule{0.2em}{0ex}}x\right]\phantom{\rule{0.2em}{0ex}}|}_{\pi \text{/}4}^{\pi }\hfill \\ & =\left(\sqrt{2}-1\right)+\left(1+\sqrt{2}\right)=2\sqrt{2}.\hfill \end{array}$

The area of the region is $2\sqrt{2}$ units 2 .

If R is the region between the graphs of the functions $f\left(x\right)=\text{sin}\phantom{\rule{0.2em}{0ex}}x$ and $g\left(x\right)=\text{cos}\phantom{\rule{0.2em}{0ex}}x$ over the interval $\left[\pi \text{/}2,2\pi \right],$ find the area of region $R.$

$2+2\sqrt{2}$ units 2

## Finding the area of a complex region

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

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

Over the interval $\left[0,1\right],$ the region is bounded above by $f\left(x\right)={x}^{2}$ and below by the x -axis, so we have

${A}_{1}={\int }_{0}^{1}{x}^{2}dx={\frac{{x}^{3}}{3}\phantom{\rule{0.2em}{0ex}}|}_{0}^{1}=\frac{1}{3}.$

Over the interval $\left[1,2\right],$ the region is bounded above by $g\left(x\right)=2-x$ and below by the $x\text{-axis,}$ so we have

${A}_{2}={\int }_{1}^{2}\left(2-x\right)dx={\left[2x-\frac{{x}^{2}}{2}\right]\phantom{\rule{0.2em}{0ex}}|}_{1}^{2}=\frac{1}{2}.$

Adding these areas together, we obtain

$A={A}_{1}+{A}_{2}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}.$

The area of the region is $5\text{/}6$ units 2 .

Good morning,,, how are you
d/dx{1/y - lny + X^3.Y^5}
How to identify domain and range
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
sorry
Dr
Dr
:(
Shun
was up
Dr
hello
is it chatting app?.. I do not see any calculus here. lol
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
by using integration product formula
Roha
find derivative f(x)=1/x
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha