What is the distance a particle has moved if I know the work done

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Homework Help Overview

The discussion revolves around determining the distance a particle has traveled given the work done, with the force expressed as an integral involving a variable upper limit. The problem is situated within the context of calculus, specifically dealing with integrals and their evaluation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenges of evaluating the integral with a variable upper limit, questioning the notation and its clarity. Some express frustration over the need to guess values for the upper limit, while others explore the implications of changing variable names to avoid confusion.

Discussion Status

Several participants have shared their attempts at evaluating the integral and have noted the upper limit they found through trial and error. There is an ongoing exploration of the process involved in reaching the solution, with some suggesting that a clearer notation could facilitate understanding. Guidance has been offered regarding the evaluation of the integral and the substitution of limits.

Contextual Notes

Participants mention the potential confusion caused by the notation used in the problem, particularly the double use of the variable 'x'. There is also a reference to the total work done being set at 108, which is central to the problem but not fully resolved within the discussion.

Psinter
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Homework Statement


I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

(x^2+2x) is the force in pounds that acts on the particle


Homework Equations


\int_{0}^{x^2+2}(x^2+2x)dx = 108


The Attempt at a Solution


I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.
 
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What is the integral of x2 + 2x?
 
alanlu said:
What is the integral of x2 + 2x?
Its \frac{1}{3}x^3+x^2
 
Psinter said:

Homework Statement


I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

(x^2+2x) is the force in pounds that acts on the particle


Homework Equations


\int_{0}^{x^2+2}(x^2+2x)dx = 108


The Attempt at a Solution


I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it.

Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write \int_{0}^{x^2+2}(x'^2+2x')dx' = 108, which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
 
Ray Vickson said:
Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write \int_{0}^{x^2+2}(x'^2+2x')dx' = 108, which makes it clear what is the integration variable (x') and what is the unknown (x).

RGV
I see what you did there, but how does that makes it easier to solve? I still don't know what to do. :frown:

Rewinding everything I know:

The force exerted on the particle: (x'^2+2x')

The initial point of the particle: 0

The distance traveled by the particle (that's my upper limit) given by: (x^2+2)

And the total work done by the particle after traveling the distance I'm looking for: 108

Still not helps writing all that. How can I find the upper limit value or the x value to plug it in and get 108 as answer?
 
Last edited:
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
 
alanlu said:
I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

Try plugging in the limits of integration as if you are evaluating the integral.
I kept trying putting in random values and found that the upper value must be 6. In other words:
\int_{0}^{6}(x'^2+2x')dx' = 108,
However, isn't there a process to reach that? I did it by brute force.
 
Psinter said:
I kept trying putting in random values and found that the upper value must be 6. In other words:
\int_{0}^{6}(x'^2+2x')dx' = 108,
However, isn't there a process to reach that? I did it by brute force.

You have already evaluated the indefinite integral F(x') = \int(x'^2 + 2x')\, dx'. Using your formula for F, what would be \int_0^a (x'^2 + 2x') \,dx'? What is preventing you from substituting in a = x^2 + 2? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
 
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
 
  • #10
Ray Vickson said:
You have already evaluated the indefinite integral F(x') = \int(x'^2 + 2x')\, dx'. Using your formula for F, what would be \int_0^a (x'^2 + 2x') \,dx'? What is preventing you from substituting in a = x^2 + 2? Mind you, the final result will not be pretty, but it gives you a place to start.

RGV
It gives me x = 2, but I wonder if there is a rule or something for me to find that the value was 6 instead of doing it by guessing like I did.

Drood said:
Here's what I did, perhaps you may find the process of some use.

1) I computed the integral and applied the limits.
2) I know that the integral with the limit substitutions = 108
3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
4) The only applicable solution I found was x=2
5) The upper limit with the x=2 substitution is indeed equal to 6.

Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

Hope that helps,

Drood
Yes please, an image would help a lot. Thanks. Its just that its weird to have to solve it by guessing, but you seem to have some process there.
 
Last edited:
  • #11
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


IMAG0630.jpg


Hope that helps,

Drood
 
  • #12
Drood said:
I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.View attachment 44430

Hope that helps,

Drood
Oh, I understand what you did there. Thank you very much. I feel dumb now because you make it look so easy and it took me so long.
 

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