# 6.1 Areas between curves  (Page 2/4)

 Page 2 / 4

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 .

find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
write something lmit
find the integral of tan tanxdx
-ln|cosx| + C
Jug
discuss continuity of x-[x] at [ _1 1]
Given that u = tan–¹(y/x), show that d²u/dx² + d²u/dy²=0
find the limiting value of 5n-3÷2n-7
Use the first principal to solve the following questions 5x-1
175000/9*100-100+164294/9*100-100*4
mode of (x+4) is equal to 10..graph it how?
66
ram
6
ram
6
Cajab
what is domain in calculus
nelson
integrals of 1/6-6x-5x²
derivative of (-x^3+1)%x^2
(-x^5+x^2)/100
(-5x^4+2x)/100
oh sorry it's (-x^3+1)÷x^2
Misha
-5x^4+2x
sorry I didn't understan A with that symbol
find the derivative of the following y=4^e5x y=Cos^2 y=x^inx , x>0 y= 1+x^2/1-x^2 y=Sin ^2 3x + Cos^2 3x please guys I need answer and solutions
differentiate y=(3x-2)^2(2x^2+5) and simplify the result
Ga
72x³-72x²+106x-60
okhiria
y= (2x^2+5)(3x+9)^2
lemmor
solve for dy/dx of y= 8x^3+5x^2-x+5
192x^2+50x-1
Daniel
are you sure? my answer is 24x^2+10x-1 but I'm not sure about my answer .. what do you think?
Ga
24x²+10x-1
Ga
yes
ok ok hehe thanks nice dp ekko hahaha
Ga
hahaha 😂❤️❤️❤️ welcome bro ❤️
Ga
y= (2x^2+5)(3x+9)^2
lemmor
can i join?
Fernando
yes of course
Jug
can anyone teach me integral calculus?
Jug
it's just the opposite of differential calculus
yhin
of coursr
okhiria
but i think, it's more complicated than calculus 1
Jug
Hello can someone help me with calculus one...
Jainaba
find the derivative of y= (2x+3)raise to 2 sorry I didn't know how to put the raise correctly
8x+12
Dhruv
8x+3
okhiria
d the derivative of y= e raised to power x
okhiria