hi. i am having trouble forming a logical chain of reasoning to prove this fact. intuitively i can see it true but i want to prove it rigorously.
from the hypothesis i know since f is integrable that for all e there exists a d such that |Rn - V| < e when |P| < d where Rn is the partial sum of a Riemann sum, V is the limit of these partial sums, and P is a partition of the interval [a,b]. i am trying to prove that there exists some c in the interval [a,b] such that integral (from a to c) f dx = 1/2 integral (from a to b) f dx.
i tried saying that since f is integrable, so is f/2 and it follows that the limit of the Riemann sums of f/2 approaches V/2. now i try to show that there exists a c such that the riemann sums 1/2 (x_i+1 - x_i)f(xi*) from (a to b) = (x_i+1 - x_i) f(xi*) (from a to c) but i am stuck here and can't seem to convince myself of this step.
is this the right way to prove this statement, and if it's not can someone give me some pointers in the right direction? thanks.
thanks for your reply! F(a) = 0 and F(b) = V which is the value of the integral. and using the intermediate value theorem, since F is continuous there exists c in [a,b] such that F(c) = V/2 since V/2 is in [o, V] and with this we are finished.
is the reason why F is continuous because F is differentiable by the fundamental theorem of calculus? other than that small point i understand the solution and can't believe i was mulling over this for quite a while. thanks!
if F(a) = F(b) there are 2 cases. if a = b then its the same point and the result follows. when a is not equal to b this is the part i am having trouble on. it seems to me that if you draw a sine curve for instance such that the positive peak cancels out with its negative peak then the integral from a to b is simply 0. then in order for the result to hold we would have to choose c to equal b on the interval [a,b]? thus in general we just need to choose a c such that the positive and negative portions cancel eachother out.
i am not sure how to turn this into a rigorous argument though. the part about choosing a c seems very handwavy to me. i thought about arguing that F(a) </= F(c) </= F(b) for at least 1 c in the interval but i am not sure if this is valid or not. if it is, then since F(a) = F(b)=0 we get F(c) = 0.