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Area using integration

paulmdrdo

Active member
May 13, 2013
386
1. find the area bounded by:
a. $\displaystyle y=x^3-3x^2-x+3$ and the x-axis for x = -1 to x = 2.

b. $y=x^3$ and $\displaystyle y=\sqrt[3]{x}$ in the first quadrant

c. $x=y^2$ and $y=x-2$

what are the steps in solving these? please help! thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

a) I would begin by finding the roots of the given polynomial (it factors nicely) to determine the intervals for which it is positive and for which it is negative. Why do we need to know this?
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

if we factor the polynomial we will get $(x-3)(x^2-1)$ so the roots of the polynomial are $x = 3$, $x=\sqrt{1}$. but i don't know the answer to your question "Why do we need to know this?" i'm hoping that you can help me. i'm just beginning to learn calculus by the way.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

If we fully factor the polynomial, we find:

\(\displaystyle f(x)=x^3-3x^2-x+3=x^2(x-3)-(x-3)=\left(x^2-1 \right)(x-3)=(x+1)(x-1)(x-3)\)

Hence we see the roots are:

\(\displaystyle x=-1,1,3\)

Because the roots are all of multiplicity 1, which simply means they are not repeated, we know the sign of $f(x)$ will alternate across the roots, so it suffices to test one interval, so let's choose $x=0$ from the interval $(-1,1)$ and we find $f(0)=3>0$.

So, we may conclude the polynomial is negative on $(-\infty,-1)$, positive on $(-1,1)$, negative on $(1,3)$ and positive on $(3,\infty)$.

Next, let's sketch the area to be found:

paulmdrdo1.jpg

Do you see now that the area under the $x$-axis will require us to integrate the negative of $f(x)$ on that interval?

When we are computing the area between two curves (and this is what we are doing if we observe that the $x$-axis is $y=0$), then we need to use as our integrand the magnitude of the difference between the two curves:

\(\displaystyle A=\int_a^b \left|f(x)-g(x) \right|\,dx\)

So, we need to know on what intervals $f(x)>g(x)$, since by definition:

\(\displaystyle |u|=u\) when $0\le u$ and $|u|=-u$ when $u<0$.

So, we see that on $(-1,1)$ we have $0<f(x)$ and on $(1,2)$ we have $f(x)<0$, hence the area can be found with:

\(\displaystyle A=\int_{-1}^2\left|f(x)-0 \right|\,dx=\int_{-1}^2\left|f(x) \right|\,dx=\int_{-1}^1 f(x)\,dx+\int_1^2 -f(x)\,dx=\int_{-1}^1 f(x)\,dx-\int_1^2 f(x)\,dx\)
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

thanks mark. but I'm still struggling to understand the entirety of your answer. how did you know the graph of that function. did you just punch it in a graphing calculator? i want to know how to draw that without using graphing calculator. what do you mean by "the sign of f(x) will alternate across the roots? thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

While I did use a graphing utility to obtain the plot of the function, it was not necessary, only easier to use to generate an image I could then attach. We really only need to know the sign of the function, so we know on which interval(s) to use the function itself as the integrand and on which interval(s) to use the negative of the function as the integrand.

When I say the sign of the function will alternate across its roots, this simply means that the sign of the function will change across each root.
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

why do we need to know the intervals which the function is positive and negative?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

Because on those intervals in which the function is negative, integrating will return a negative value. so we need to negate the function on those intervals to return a positive value. If we simply integrate the given function over the entire interval, then we will get the area above the $x$-axis minus the area below. But, we want to add the two areas to get the total area.
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

how do we know that the sign of f(x) will alternate across each roots without looking at the graph? and also how can we determine that there would be an "alternating of the sign situation" of the f(x) in the problem?
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
Re: area using integration

how do we know that the sign of f(x) will alternate across each roots without looking at the graph?

You don't need to... Let's say f(x) has a root at x= -2. Then simply calculate f(x) for x+z and x-z for a very small z. You will then find that, if f(x) changes sign around the root x=-2, then f(x+z) and f(x-z) will also have different sign... ;)
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

my answer to a would be 5.75 sq. units is it correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

how do we know that the sign of f(x) will alternate across each roots without looking at the graph? and also how can we determine that there would be an "alternating of the sign situation" of the f(x) in the problem?
When a root is of odd multiplicity, then we know the function will change sign across this root. Think of the graphs of $y=x^{2n}$ vs. $y=x^{2n+1}$ where $n\in\mathbb{N}$ as an example.

my answer to a would be 5.75 sq. units is it correct?
Yes, that is correct. For part b) I would begin by determining the limits of integration and which function is greater on that interval. What do you find?
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

to determine the limits of integration i would set both function equal to one another.

$\displaystyle x^3=\sqrt[3]{x}$ then we will have $\sqrt[3]{x}(\sqrt[3]{x^2}-1)$

the limits of integration would be x=0;x=1

but i don't know how to determine which function is greater.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

One way is to pick a test value for $x$ on the interval $(0,1)$. I would try \(\displaystyle x=\frac{1}{2}\).
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

One way is to pick a test value for $x$ on the interval $(0,1)$. I would try \(\displaystyle x=\frac{1}{2}\).
where will I plug that test value?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

where will I plug that test value?
Into both of the bounding functions. Whichever returns the greater value will be the greater function on that interval.
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

$\sqrt[3]{x}$ is greater.

calculating

$\displaystyle \int_{0}^1{\sqrt[3]{x}-x^3}dx$

my answer would be 1/2 or .50 sq. units did i get it right?

 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

Yes, that's correct. For part c) I recommend switching the variables...
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

Yes, that's correct. For part c) I recommend switching the variables...
what do you mean by switching the variables?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

Let $x$ become $y$ and $y$ become $x$ so that you have:

\(\displaystyle y=x^2\)

\(\displaystyle y=x+2\) (this one has been rearranged from \(\displaystyle x=y-2\))

You don't have to switch them like this, but you can and it may make it easier for you to determine the area.
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

the limits of integration are x=2;x=-1

we now have

$\displaystyle\int_{-1}^2{(x+2-x^2)dx}$

then arranging the integrand we have

$\displaystyle\int_{-1}^2{(-x^2+x+2)dx}$

the answer is 6.5 sq units?

for educational purposes, can we solve this using other method? can you show me how would that work?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

No, 6.5 is not correct. Can you show your work?

Another way would be:

\(\displaystyle A=\int_{-1}^2 (1+x)(2-x)\,dx\)

Use the substitution:

\(\displaystyle u=1+x\,\therefore\,du=dx\)

and we have:

\(\displaystyle A=\int_0^3 u(3-u)\,du=\int_0^3 3u-u^2\,du\)

This is easier to evaluate because the lower limit is $0$.
 

paulmdrdo

Active member
May 13, 2013
386
Re: area using integration

this is how i solve part c)



$\displaystyle\int_{-1}^2 (-x^2+x+2)dx$

taking the definite integral of the terms individually

$\displaystyle\int_{-1}^2 -x^2dx+\int_{-1}^2 xdx +\int_{-1}^2 2dx$

i will have 9/2 sq. units. is this correct?


in your solution above why did change the limits of integration?




 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: area using integration

this is how i solve part c)



$\displaystyle\int_{-1}^2 (-x^2+x+2)dx$

taking the definite integral of the terms individually

$\displaystyle\int_{-1}^2 -x^2dx+\int_{-1}^2 xdx +\int_{-1}^2 2dx$

i will have 9/2 sq. units. is this correct?


in your solution above why did change the limits of integration?




Yes, 9/2 is correct.

When you are evaluating a definite integral, and you change the variable, you have to also change the limits so that they are in terms of the new variable. I used:

\(\displaystyle u=1+x\)

The original lower limit was -1, and so:

\(\displaystyle u(-1)=1+(-1)=0\)

So the new lower limit is 0.

The original upper limit was 2, and so:

\(\displaystyle u(2)=1+(2)=3\)

So the new upper limit is 3.