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

by Psinter
Tags: distance, moved, particle, work
 P: 97 1. The problem statement, all variables and given/known data 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 2. Relevant equations $\int_{0}^{x^2+2}(x^2+2x)dx = 108$ 3. 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.
 P: 65 What is the integral of x2 + 2x?
P: 97
 Quote by alanlu What is the integral of x2 + 2x?
Its $\frac{1}{3}x^3+x^2$

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P: 5,086
What is the distance a particle has moved if I know the work done

 Quote by Psinter 1. The problem statement, all variables and given/known data 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 2. Relevant equations $\int_{0}^{x^2+2}(x^2+2x)dx = 108$ 3. 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
P: 97
 Quote by Ray Vickson 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.

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?
 P: 65 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.
P: 97
 Quote by alanlu 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.
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P: 5,086
 Quote by Psinter 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
 P: 8 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
P: 97
 Quote by Ray Vickson 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.

 Quote by Drood 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.
 P: 8 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. Hope that helps, Drood
P: 97
 Quote by Drood 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. 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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