Finding the area of a region

[CatchThis2]

I need to find the area of a region when y=e^4x and when y=e^7x and x=1. If anyone has the time I would really appreciate them looking at this for me.

 

[mmm4444bot]

Are you talking about the area between e^(4x) and e^(7x) from x = 0 through x = 1 ?

The image below shows e^(7x) in red and e^(4x) in green from x = 0 through x = 0.1 .

[attachment=0:wf23zspr]exp(Xx).JPG[/attachment:wf23zspr]

To calculate the area between the red and green graphs, from the point (0, 1) where they meet, all the way out to the vertical boundary at x = 1, we calculate the area under each curve separately. Then, subtract the smaller from the larger.

Do you understand that the two symbols below represent these two areas?

$\displaystyle \int^1_0 e^{7x} \ dx$

$\displaystyle \int^1_0 e^{4x} \ dx$

Please clarify your question, and try to make some statements about what you've learned, so far, about any of this stuff, so that we might determine where to begin helping you.

Cheers ~ Mark

MY EDIT: Removed incorrect statement.

 

[swaroop]

Physical meaning of integral is area under that function. So we need to find area bound by e^7x, e^4x and x=1.
So from the plot in the above post
Area=Inte(e^7x-e^4x) with limits from 0 to 1.
So Area= (e^7/7- e^4/4)-(1/7-1/4)

p.s- Sorry I am not sure abt printing integ and fraction prop.

 

[JuicyBurger]

Yes, swaroop is correct, although maybe I can make it a bit more readable.

$\displaystyle \int_{0}^{1}(e^{7x}-e^{4x})dx = [\frac{1}{7}e^{7x}-\frac{1}{4}e^{4x}]_{0}^{1}$

$\displaystyle [\frac{1}{7}e^{7}-\frac{1}{4}e^{4}] - [\frac{1}{7}e^{0}-\frac{1}{4}e^{0}]$

Finishing this up should yield the correct answer.

Hope that helps!

 

[CatchThis2]

Not quite sure how to finish that off. I don't have a good calculator on me to finish off the work you have. Could someone finish off this answer for me. Thanks alot!

 

[JuicyBurger]

What do you mean you are not sure how to finish that off?

Maybe you should try to figure that one out on your own and not expect to be spoon-fed the answers without even the slightest effort on your part.

 

[mmm4444bot]

In the words of E. M. Forster, "Spoon feeding, in the long run, teaches us nothing but the shape of the spoon."

BTW: The calculator bundled with Windows is more than sufficient.

 

[CatchThis2]

Not looking to be "spoon fed" just trying to learn how to do my homework based on the work guys have submitted to me. I got work with answers earlier and was able to go back through the problem and understand how they got it so it was beneficial to me. I am not very good at calculus or math in general but just trying to get through this last college math class. Hope you understand!

 

[CatchThis2]

Another find the area of the region question: y=4x^2 , y=(x^2+5)

So far I set the equations equal to another and solved to get 3x^2-5 then I factored that out and got 1(3x^2-5)

Next I set 1=to 0
Next I set 3x^2-5= to 0 and solved and got 5/6= to 0

That's where I am now stuck.

 

[JuicyBurger]

First question is, why did you factor out a 1, and how did you set $\displaystyle 1=0$!?!?!?

When you set $\displaystyle 3x^2-5 = 0$ how did you get $\displaystyle 5/6$?? On a side note, you said that $\displaystyle 5/6 = 0$, when you mean that $\displaystyle x = 5/6$

$\displaystyle 3x^2 = 5$
$\displaystyle x^2 = 5/3$
$\displaystyle x = \pm\sqrt{\frac{5}{3}}$

This is where the graphs intersect each other, and therefore they will be your limits of integration.

Hope this helps!

 

[CatchThis2]

So your saying I just plug 5/3 into both equation and solve and I should get the answer? I am not very sure.

 

[JuicyBurger]

So your saying I just plug 5/3 into both equation and solve and I should get the answer? I am not very sure.

What I am saying is that after you do the work I described you should get $\displaystyle x = \pm\sqrt{5/3}$ (NOT just 5/3, see the square root?).

This is the result of setting the two equations equal to each other and yields the x-coordinates where they INTERSECT. Therefore when you do your integration you should set the smaller of the two ($\displaystyle -\sqrt{5/3}$)as your lower limit, and the larger of the two($\displaystyle \sqrt{5/3}$) as your upper limit.

You still need to find out what function(s) to integrate, much in the same way we did the first problem in this thread. Try to figure this function out and you should have everything you need.

 

[CatchThis2]

For the first area problem I got (1/4)e^3 + (1/4) . Is this the answer you got.


I see how you got +-sqrt (5/3) silly error on my part. Thanks for picking it up!

 

[JuicyBurger]

$\displaystyle [\frac{1}{7}e^{7}-\frac{1}{4}e^{4}] - [\frac{1}{7}e^{0}-\frac{1}{4}e^{0}]$

This is the answer for the first question.

Do you know how to find the function to integrate for this new problem? You find it the same way as last time, by subtracting the one on bottom from the one on top.

Also, it helps to make a new thread for each new question.

Hope this helps!

 

[CatchThis2]

I see how you got that answer. Since e^7 and e^4 are not like terms you can't simplify them. I kept trying to subtract and got the wrong answer. Thanks for taking the time to explain that and about posting my different questions as new topics.

 

[BigGlenntheHeavy]

$\displaystyle Second \ One:$

$\displaystyle A \ = \ 2\int_{0}^{\sqrt(5/3)}[5-3x^{2}]dx \ = \ \frac{20\sqrt(15)}{9} \ sq. \ units$

 

[CatchThis2]

Thanks for showing the work and how you arrived at your answer. I know it takes alot of time and patients to plug everything in. Seeing all the work helps out very much Thank you very much as I really appreciate your time!

 

[swaroop]

e^7=1096.63
e^7/7=156.66
e^4=54.598
e^4/4=13.65
[e^7/7-e^4/4]=143.01
[1/7-1/4]=-0.107
Final answer is 143.01-(-0.107)=143.117
@catchthis2- Please try to put more efforts. Sorry if I am offensive. Hope the answer is correct