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$F(x) = \int_{x-1}^{x+1}f(t)dt$ for x an element of the reals.

Show that $F$ is differentiable on Reals, and compute $F^\prime$.

I am unsure about how to showing $F$ is differentiable. I know that I need to use the fundamental theorem of calculus, but can someone please explain how to do so?

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3 Answers 3

up vote 1 down vote accepted

You can simply use definition of the derivative.

You have $$F(x)=\int_{x-1}^{x+1} f(t) dt.$$

$$F(x+h)-F(x)=\int_{x+h-1}^{x+h+1} f(t) dt-\int_{x-1}^{x+1} f(t) dt= \int_{x+1}^{x+h+1} f(t) dt - \int_{x-1}^{x+h-1} f(t) dt$$

$$\frac{F(x+h)-F(x)}{h}=\frac{\int_{x+1}^{x+h+1} f(t) dt}{h} - \frac{\int_{x-1}^{x+h-1} f(t) dt}h$$

$$ \min_{c\in_\langle x+1,x+1+h\rangle} f(c)-\max_{c\in_\langle x-1,x-1+h\rangle} f(c) \le \frac{F(x+h)-F(x)}{h} \le \max_{c\in_\langle x+1,x+1+h\rangle} f(c)-\min_{c\in_\langle x-1,x-1+h\rangle} f(c)$$ (From the continuity we know that the minima and maxima exists.)

Now (also from the continuity) both expression converge to $f(x+1)-f(x-1)$.

However, you might want to have a look at a general version of this http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

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If $f$ is not continuous, then $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ need not be differentiable at the points of discontinuity of $f$.

For instance I believe the following is a counter-example:

$ f(x) = \begin{cases} 0 & x \le 3 \\ 1 & x \gt 3 \end{cases}$

I believe we can show that $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ is not differentiable $2$.

However, it is a well known theorem that at any point of continuity $c$ of $f$, the function $G(x) = \int_{a}^{x} f(t) \text{d}t$ is differentiable and $G'(c) = f(c)$.

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If you haven't done any measure theory , a simple answer would be:

If f is continuous then it has a primitive (the integral is supposed to be rieman one) If 0 belongs to the domain of f, let us then call G(x)=integral from 0 to x f(t)dt such a primitive function wich is differentiable (as f is its derivative and f is continuous).

F(x)=G(x+1)-G(x-1). F'(x)=(x+1)'f(x+1)-(x-1)'f(x-1).

Thus

F'(x)=f(x+1)-f(x-1).

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I don't know aything about it being a primitive –  mary Apr 25 '11 at 4:50
    
Can you suggest another way of doing this problem, and demonstrate it so I can see how to replicate it for others? –  mary Apr 25 '11 at 4:50
    
a primitive function is simply the opposite operator to derivation. the primitive of a function f is the one whose derivative is f. When f is continuous the primitive does exist and it is the integral + a constant. so here if say f is defined in 0 and continuous on the whole set of reals. you can take G(x) to be the integral from 0 to x of f(t)dt. –  El Moro Apr 25 '11 at 5:13
    
This sounds circular to me. After all, why is it true that if $f$ is continuous then it has a primitive? Sure, we can define a function $G$ by $G(x) = \int_0^x f(t)\,dt$... but how do we know that $G'(x) = f(x)$? I mean, that's what we're trying to prove! –  Jesse Madnick Apr 25 '11 at 5:22
1  
Ah, I see we're assuming the Fundamental Theorem of Calculus is true... mm, okay then. –  Jesse Madnick Apr 25 '11 at 5:27

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