Fundamental theorem of calculus

[mglaros]

Homework Statement

Does the function F(x)=int(sin(1/t)dt,0,x) (integral of sin(1/t) with lower limit 0 to upper limit x) have a derivative at x=0?

Homework Equations

The Attempt at a Solution

I was thinking that F(x) shouldn't have a derivative at x=0 because the integrand isn't even continuous at 0. I tried making this more explicit through using the definition of the derivative along with the convention that F(0)=0 because sin(1/t) is bounded.

Any suggestions? Is my reasoning correct?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Dick]

This one's kind of tricky. I had to think about it for a while. F(x) actually IS differentiable at 0. Try to prove this by showing -x^2<=|F(x)|<x^2. Big hint: integrate t*cos(1/t) by parts.

 

[mglaros]

okay, once I have shown -x^2<=|F(x)|< x^2 then taking the limit as x tends to zero shows that F(x) is zero. My question now is how does this give us any information on the differentiability of F(x)? Is there something that I am missing?

Thank you for your time

 

[snipez90]

$$|F'(0)| = \lim_{h \rightarrow 0}\left|\frac{F(h) - F(0)}{h}\right| = \lim_{h \rightarrow 0}\frac{|F(h)|}{|h|} \leq \lim_{h \rightarrow 0}\frac{|h|^2}{|h|} = 0.$$

Hence F'(0) = 0.

 

[mglaros]

wow, I don't see how I overlooked that haha. I really need to get some sleep. Thank you both for your help!