# If F(x)=$\int_0^xf(x)dx$ for $x\in[0,3]$, find $F$. Find $F'(x)$ at all points where $F'$ exists.

Let $f$ is given by $$f(x)=\begin{cases} x,~~~ 0\leq x \leq 1\\ 1, ~~~1\leq x\leq 2 \\ x, ~~~2<x\leq 3 \end{cases}$$ If F(x)=$\int_0^xf(x)dx$ for $x\in[0,3]$, find $F$. Find $F'(x)$ at all points where $F'$ exists.

Attempt:

As f has only one point of discontinuity at $x=2$, so f is Riemann integrable on $[0,3]$. Then $$F(x)=\begin{cases} \int_0^x x\, dx,~~~ 0\leq x \leq 1\\ \int_0^x 1\, dx, ~~~1<x\leq 2 \\ \int_0^x x\, dx, ~~~2<x\leq 3 \end{cases}=\begin{cases} \frac{x^2}{2},~~~ 0\leq x \leq 1\\ x-1/2, ~~~1<x\leq 2 \\ x^2/2-1/2, ~~~2<x\leq 3 \end{cases}$$

As $f$ is not continuous, then by Fundamental Theorem of Calculus, $F$ is not differentiable everywhere on $[0, 3]$. If the above correct, please help for remaining part.

• While the final result of your computation of $F(x)$ is correct, the intermediate step should be: $$F(x)=\begin{cases} \int_0^x x\, dx,~~~ 0\leq x \leq 1\\ \int_0^1 x\,dx +\int_1^x 1\, dx, ~~~1<x\leq 2 \\ \int_0^1 x\, dx +\int_1^2 1\, dx +\int_2^x x\, dx, ~~~2<x\leq 3 \end{cases}$$ – Torsten Schoeneberg Mar 26 '18 at 5:58

Your computations for $F$ are correct. Now look at the intervals $[0,1),(1,2)$ and $(2,3]$. By the Fundamental Theorem of Calculus, $F$ is differentiable on these intervals. It is your turn to investigate the differentiability of $F$ at the points $1$ and $2$.