Hi everyone,
First of all, great forum! Hope someone out there can help me : )
Here`s the problem:
Let $\displaystyle f(x) = sin( \frac{1}{x}) for x \neq 0, f(0) = 1$
Calculate the derivative of $\displaystyle F(x) = \int_0^x f(t)dt$
Hi everyone,
First of all, great forum! Hope someone out there can help me : )
Here`s the problem:
Let $\displaystyle f(x) = sin( \frac{1}{x}) for x \neq 0, f(0) = 1$
Calculate the derivative of $\displaystyle F(x) = \int_0^x f(t)dt$
You mean the answer is just: $\displaystyle sin( \frac{1}{x} )$?
(That`s way too easy)
Anyway, thanks a lot for the quick reply : )
Ok, one more question, so I don`t open a new thread:
How can I proof that f(x) is integrable over every interval [a,b]?
EDIT:
I guess this helps a lot xD
"If f(x) is bounded on [a,b] and has finitely many points of discontinuity on [a,b],
then f is integrable on [a,b]."
You are going to need to know how to calculate a definite integral to solve this problem, and if you are doing calculus, you really ought to know how to calculate a definite integral! This problem doesn't make sense without it.
EDIT: Sorry, as I wasn't thinking right when I gave this answer, Vince's answer below is better.
so in your case the answer is what felper said. I'm very sure.
On-Edit: Everything on the right hand side in the equaiton above is 0 except the second term which is your answer. there is no need to worry about the limits because you're differentiating with respect to x not t. Had the integrand been a function of both x and t, then you would have had to worry about the limits.