# 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. 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 $f\left(x\right)=g\left(x\right),$ 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 region between two curves can be broken into two sub-regions.

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 $R.$ 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 $f\left(x\right)=g\left(x\right)$ 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 .

#### Questions & Answers

find the domain and range of f(x)= 4x-7/x²-6x+8
Nick Reply
find the range of f(x)=(x+1)(x+4)
Jane Reply
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
Harrieta Reply
d/dx{1/y - lny + X^3.Y^5}
mogomotsi Reply
How to identify domain and range
Umar Reply
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
Adri
sorry
Dr
hi adri ana
Dr
:(
Shun
was up
Dr
hello
Adarsh
is it chatting app?.. I do not see any calculus here. lol
Adarsh
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
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
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
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)
Inembo Reply
what is calculus?
BISWAJIT Reply
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?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
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
Adri
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).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
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
Adri
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
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
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

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