Does the Integral Converge for p<0 in int[exp(px)]dx?

  • Thread starter Thread starter dan
  • Start date Start date
  • Tags Tags
    Value
AI Thread Summary
The integral of exp(px) from 0 to infinity converges only for p<0, where its value is -1/p. For p=0, the integrand becomes constant, leading to an unbounded result. When p>0, the integrand increases without bound, resulting in divergence. As p approaches 0 from the negative side, the integral approaches infinity due to the behavior of exp(px). Thus, the integral does not converge for p≥0, confirming that convergence is strictly limited to p<0.
dan
Please, can anyone help me here with this problem??

Q) Determine for what values of

element p is a member of set R, int[exp(px)]dx ; where
upper lim=infinity , lower lim=0

converges and find its value in those cases?

so far:

I got the anti-derivative of the integrand is exp(px)/p.

For the integral to exist, you must have p<0, in which case the
value is -1/p.

The integrand increases without bound for p>0, and for p=0, it is constant, again leading to an unbounded result for the value.

This is where I'm lost, how do you find the value for p<0?

Can anyone verify what I have done and help me solve this problem.

Thanks for your help.
 
Last edited by a moderator:
Physics news on Phys.org
This is where I'm lost, how do you find the value for p<0?[/QUOTE}

Uhhh, didn't you just say:
For the integral to exist, you must have p<0, in which case the
value is -1/p.
?

In any case, the problem does not ask you to find the value of the integral- it only asks you to determine for what VALUES OF p the integral exists- you've already done that! (0f course, to show that your answer is correct, it is good to actually display the value: again, you've already done that!)

If I were really hard-nosed (and I AM!) I might point out that there is one error here: the anti-derivative of exp(px)= exp(px)/p only for p NOT equal to 0.

If p= 0, what is exp(px)? What is it's anti-derivative? Can that be evaluated between 0 and infinity?
 


To find the value for p<0, we can use the limit definition of the integral. As p approaches 0 from the negative side, the integral will approach infinity. This is because the integrand, exp(px), will approach 1 as p approaches 0, but the interval of integration is from 0 to infinity, so the area under the curve will continue to increase without bound.

Therefore, we can conclude that for p<0, the integral does not converge and does not have a finite value. This is because the function is increasing without bound and the interval of integration is infinite.

In summary, for the integral int[exp(px)]dx to converge, the value of p must be strictly less than 0. In those cases, the value of the integral is -1/p. For p≥0, the integral does not converge.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top