0
$\begingroup$

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.

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

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 $2$.

$\endgroup$
3
  • $\begingroup$ Thank you. looking at the intervals [0,1),(1,2) and (2,3] we see that f is continuous at each of the intervals. Please explore "By the Fundamental Theorem of Calculus, F is differentiable on these intervals". $\endgroup$ – user1942348 Mar 26 '18 at 6:53
  • $\begingroup$ Ooops , a typo, my fault ! $\endgroup$ – Fred Mar 26 '18 at 6:58
  • $\begingroup$ Its ok I understand. Please explore "By the Fundamental Theorem of Calculus, F is differentiable on these intervals" Are you want to mention as f in continuous on [0,1),(1,2) and (2,3], then by Fundamental Theorem of Calculus, F is differentiable on those intervals. $\endgroup$ – user1942348 Mar 26 '18 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.