From the fundamental theorem of calculus we have \begin{align} F(x)&=\int_a^x f(t) \, dt \tag 1\\ \frac{dF(x)}{dx}&=f(x) \tag 2 \end{align} If I integrate equation (2) I get \begin{align} \int_a^xdF&=\int_a^xf(t) \, dt \iff \\ F(x)-F(a)&=\int_a^xf(t) \, dt \iff \\ F(x)&=F(a)+\int_a^xf(t)\, dt \tag 3 \end{align} But (3) isn't the same as equation (1), shouldn't it be that?

  • 2
    $\begingroup$ $F(a)=0$, so there's no contradiction $\endgroup$ – user281392 Jul 13 '17 at 16:37

Since $F(x)$ is defined as a definite integral starting at $a$, then we have that $F(a)=0$, so there's no problem.

By definition,

$$F(a)=\int_a^a f(t)\,dt = 0$$

Does that help?


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.