2
$\begingroup$

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

3 Answers 3

1

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

3

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).

2

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).

  • 0
    @user I edited my answer to make it accessible this time :). hope it works for you2019-01-01