# Prove function is differentiable

• May 12th 2010, 07:16 PM
Pinkk
Prove function is differentiable
Let $f$ be continuous on the reals and let $F(x) = \int_{x-1}^{x+1}f(t)dt$. Show that $F$ is differentiable on the reals and compute $F'$.

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 $F(x) = \int_{0}^{x+1}f(t)dt - \int_{0}^{x-1}f(t)dt$ but I'm not sure how to continue from there via utilizing the FTC. Any help would be appreciated.
• May 12th 2010, 07:20 PM
Drexel28
Quote:

Originally Posted by Pinkk
Let $f$ be continuous on the reals and let $F(x) = \int_{x-1}^{x+1}f(t)dt$. Show that $F$ is differentiable on the reals and compute $F'$.

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 $F(x) = \int_{0}^{x+1}f(t)dt - \int_{0}^{x-1}f(t)dt$ but I'm not sure how to continue from there via utilizing the FTC. Any help would be appreciated.

FTC maybe overkill but clearly both of the things you broke the integral into are differentiable by the FTC and therefore is their difference.
• May 12th 2010, 07:32 PM
Pinkk
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 $F(x)=\int_{a}^{x}$ for $x\in [a,b]$. Would a proof go like this then? :

Observe that $F(x)=\int_{0}^{x+1}f(t)dt - \int_{0}^{x-1}f(t)dt$. Whether $x+1,x-1$ are positive or negative, since $f$ is continuous on the reals, each corresponding integral is differentiable and so $F$ is differentiable and since the sum of the derivatives is equal to the derivative of the sums, $F'(x) = f(x+1) - f(x-1)$. 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?
• May 12th 2010, 07:37 PM
Drexel28
Quote:

Originally Posted by Pinkk
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 $F(x)=\int_{a}^{x}$ for $x\in [a,b]$.

$a,b$ can be negative......

Quote:

Observe that $F(x)=\int_{0}^{x+1}f(t)dt - \int_{0}^{x-1}f(t)dt$. Whether $x+1,x-1$ are positive or negative, since $f$ is continuous on the reals, each corresponding integral is differentiable and so $F$ is differentiable and since the sum of the derivatives is equal to the derivative of the sums, $F'(x) = f(x+1) - f(x-1)$. Q.E.D.
Correctomundo.

Quote:

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?
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.
• May 12th 2010, 07:52 PM
Pinkk
Yeah I was trying to show that for any $a\in \mathbb{R},\lim_{x\to a} \frac{F(x)-F(a)}{x-a}=f(a+1) - f(a-1)$ but I got stuck.