Improper Integrals: Simplifying & Evaluation

[noname1]

I have a couple of integrals where i have to simplify using derive 6 and now i have to say if there are improper, if they have an infinite discontinuity or can't be evaluated.

I was wondering if someone could briefly explain what an improper is and that for example i can say that this equation have an infinite discontinuity at 1 or do i have to first solve the integral and than look at the answer?

∫(1/(x - 1)^(2/3)) from 0 to 2

 

[bblenyesi]

You've basically said the definition yourself. An improper integral is an integral in which an infinite discontinuity appears in one of the limit points, or inside the interval. There are numerous methods of solving these integrals (they constitute a series of 14 courses and seminars, each 2 hours during this semester at my university). I think that in this case what you can do is multiply the whole thing by $$x^{2/3}$$, and then test for convergence. If it converges then a conventional method will do. Then again there must be easier ways to do it, because I'm not that good at maths.

 

[noname1]

that doesn't make much sense because i have to first say if they are improper or not, second if they have a discontinuity in them. From what i understand not all improper integrals have a discontinuity in them correct?

than i was also wondering for example i can say that this equation have an infinite discontinuity at 1 or do i have to first solve the integral and than look at the answer?

∫(1/(x - 1)^(2/3)) from 0 to 2

 

[bblenyesi]

Sorry for clouding up things if I did :). You are correct, not all of them have a discontinuity in them. Your integral is an improper Riemann integral of the second kind: http://en.wikipedia.org/wiki/Improper_Riemann_integral. In the right hand corner there are some examples of improper integrals and they are explained better then I can explain them :).

So yes, it is an improper integral with an infinite discontinuity in 1, there's no need to solve it.
If you want to have a more elegant proof, a graphical solution is best for it. If you plot the integral, you will observe the existence of a vertical asymptote in the graph, so the area of the graph defined by the definite integral cannot be calculated, hence it is improper.

Hope this helps.
B.