Let be continuous on the reals and let . Show that is differentiable on the reals and compute .
So this should seem straightforward but I'm having trouble actually formalizing it simply because the lower and upper limits of the integral are both variables. I thought of saying that but I'm not sure how to continue from there via utilizing the FTC. Any help would be appreciated.
I guess I'm just hung up on notation because the upper limits of both integrals can very well be less than zero and the theorem applies to a function of the form of for . Would a proof go like this then? :
Observe that . Whether are positive or negative, since is continuous on the reals, each corresponding integral is differentiable and so is differentiable and since the sum of the derivatives is equal to the derivative of the sums, . Q.E.D.
And this does seem like overkill but we just learned the FTC so I want to make sure I don't assume too much. What would be the easier method of proof?
can be negative......
Correctomundo.Observe that . Whether are positive or negative, since is continuous on the reals, each corresponding integral is differentiable and so is differentiable and since the sum of the derivatives is equal to the derivative of the sums, . Q.E.D.
Not easier. There is just a consensus, that one should work as hard as you can in math. If you can do something without invoking an advanced theorem one should. That said, it's doable but much more work (and would be in effect mimicking the actual proof of the FTC) and so just stick with the above.And this does seem like overkill but we just learned the FTC so I want to make sure I don't assume too much. What would be the easier method of proof?