# If f is Riemann integrable and continuous, prove that F is differentiable

• Apr 22nd 2009, 12:34 PM
Last_Singularity
If f is Riemann integrable and continuous, prove that F is differentiable
Some help on this following question, please. Thanks a bunch!

Question: If $\displaystyle f$ is Riemann integrable on $\displaystyle [a,b]$ and continuous at $\displaystyle x_0$, prove that $\displaystyle F(x) = \int_{a}^{x} f(t) dt$ is differentiable at $\displaystyle x_0$ and $\displaystyle F'(x_0)=f(x_0)$. Show that if $\displaystyle f$ has a jump discontinuity at $\displaystyle x_0$, then $\displaystyle F$ is not differentiable at $\displaystyle x_0$.
• Apr 22nd 2009, 01:41 PM
Plato
I sure that you mean $\displaystyle F(x) = \int_a^x {f(t)dt}$.

Notice that $\displaystyle F(x_0 + h) - F(x_0 ) = \int_{x_0 }^{x_0 + h} {f(t)dt}$ and the length of that interval is $\displaystyle |h|$.
• Apr 22nd 2009, 01:58 PM
manjohn12
Consider the difference quotient: $\displaystyle \frac{F(x)-F(x_0)}{x-x_0} = \frac{\int_{a}^{x} f(t) \ dt- \int_{a}^{x_0} f(t) \ dt}{x-x_0} = \frac{1}{x-x_0} \int_{x_0}^{x} f(t) \ dt$.

Then you need to show that $\displaystyle \lim\limits_{x \to x_0} \frac{1}{x-x_0} \int_{x_0}^{x} f(t) \ dt = f(x_0)$.

Use the fact that $\displaystyle f(x_0) = \frac{1}{x-x_0} \int_{x_0}^{x} f(x_0) \ dt$. Also use the fact that $\displaystyle \left| \int_{a}^{b} f \right| \leq \int_{a}^{b} |f|$ provided that $\displaystyle f$ is Riemann integrable. You have to consider two cases: $\displaystyle x_0 < a$ and $\displaystyle x_0 > a$.