The second part of the fundamental theorem of calculus states that if
$$ f(x)=g'(x) $$
then
$$ \int^b_a f(x)\,dx = g(b)-g(a)$$
And so I was wondering that if you solved $\int^b_a f(x)\,dx$ using some other method like the limit definition and showed to to be equal to some $g(b)-g(a)$, then does that imply that $g'(x)=f(x)$?
This question came to me when I was solving for $\int^b_a x\,dx$ using the limit definition of the integral and I showed it to be equal to $\frac{b^2-a^2}2$, and so I was just wondering if I could do this with any integral that could be solved using alternate methods.
Best Answer
You do need some additional conditions on $f(x)$ to make it work.
Consider the function $f(x) = \begin{cases}1 & x \ \text{is an integer,} \\ 0 & x \ \text{is not an integer.}\end{cases}$
For any reals $a,b$, we have $\displaystyle\int_a^b f(x)\,dx = 0 = g(b)-g(a)$, where $g(x) = 0$.
But $g'(x) = 0 \neq f(x)$ for integer values of $x$.
If we add the condition that $f(x)$ is continuous, then your claim works.