If function is Riemann integrable on a closed interval, does it have a primitive function on the interval

calculusintegrationriemann-integration

I'm trying to solve the following theoretical question:

If function is Riemann integrable on interval [a,b], does it have a
primitive function on [a,b]?

My solution is the following:

$f(x)=\left\{\begin{matrix}
1, & x\in \left [ 0,2 \right )\\
2, & x\in \left [ 2,4 \right ]
\end{matrix}\right.$

The following function is Riemann integrable on [0,4] but a primitive function does not exist on interval [0,4]. Therefore the statement is not correct. Is it correct? If not, could you help me fix it?

Best Answer

This is false. There are many functions that are not the derivative of anything, yet they may have integrals.

In particular, no function with a simple discontinuity is a derivative. Thus, any such function that is bounded will serve as counterexample.

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