# 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 .

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).
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
What is the first fundermental theory of Calculus?
I want simple integral
for MSc chemistry... simple formulas of integration
aparna
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funny
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funny
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value of log ax cot-x cos-x
aparna
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aparna
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Hi
RIZWAN
I don't understand the formula
who's formula
funny
What is a independent variable
a variable that does not depend on another.
Andrew
solve number one step by step
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel
how to prove tan^2 x=csc^2 x tan^2 x-1?
divide by tan^2 x giving 1=csc^2 x -1/tan^2 x, rewrite as: 1=1/sin^2 x -cos^2 x/sin^2 x, multiply by sin^2 x giving: sin^2 x=1-cos^2x. rewrite as the familiar sin^2 x + cos^2x=1 QED
Barnabas
how to prove sin x - sin x cos^2 x=sin^3x?
sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x =sinx-sinx(1+sin^2x) =sinx-sinx+sin^3x =sin^3x thats proved.
Alif
how to prove tan^2 x/tan^2 x+1= sin^2 x
Rochel
Salim
what is function.
what is polynomial
Nawaz
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Alif
a term/algebraic expression raised to a non-negative integer power and a multiple of co-efficient,,,,,, T^n where n is a non-negative,,,,, 4x^2
joe
An expression in which power of all the variables are whole number . such as 2x+3 5 is also a polynomial of degree 0 and can be written as 5x^0
Nawaz
what is hyperbolic function
find volume of solid about y axis and y=x^3, x=0,y=1
3 pi/5
vector
what is the power rule
Is a rule used to find a derivative. For example the derivative of y(x)= a(x)^n is y'(x)= a*n*x^n-1.
Timothy