Recent content by ThomasTheTan

Hmmm... the only thing that bothers me about it is the assumption that f is continuous (except at c). Otherwise, it's good as far as I can tell...

Oh, good call. I didn't even think of trying anything by contradiction. I think that what you've written works perfectly! (My previous post was in regard to my proof!) Thanks!

Okay, so that proof doesn't work. Here's why: when I'm given epsilon, I haven't actually chosen a value at which the function should be continuous. That is, I define the area that is continuous based on the choice of epsilon (and, therefore, delta). For this reason, the area that I'm...

Okay, so here's a solution that I just came up with (but I'm a bit iffy about using the limit to establish continuity near a point): Lim(x->c+)[f] = L => Given epsilon > 0, there exists delta > 0 such that whenever 0<x-c<delta we have |f(x) - L| < epsilon/2. Consider an arbitrary d such...

Thanks for you quick response! Although this does provide the solution intuitively, I'm supposed to come up with a rigorous proof of this fact. Additionally, F actually is continuous at c (according to the fundamental theorem of calculus). But, I agree, the area under the curve before c...

Homework Statement Given: f is integrable on [a,b] and has a jump discontinuity at c in (a,b) (meaning that both one-sided limits exist as x approaches c from the right and from the left but that Lim(x->c-)[f] is not equal to Lim(x->c+)[f]). Show that F(x) = Integrate[f,{a,x}] is not...