In first fundamental theorem of calculus,it states if $A(x)=\int_{a}^{x}f(t)dt$ then $A'(x)=f(x)$.But in second they say $\int_{a}^{b}f(t)dt=F(b)-F(a)$,But if we put $x=b$ in the first one we get $A(b)$.Then what is the difference between these two and how do we prove $A(b)=F(b)-F(a)$?

  • $\begingroup$ If $f$ is continuous, then the first theorem's consequent is true, which in turn guarantees that the second theorem's consequent is also true.$$$$ If on the other hand $f$ is not continuous, then the first theorem is inapplicable; if $f$ is integrable and has a primitive (despite not being continuous), then the second theorem's consequent is true. p.s. There exist functions that are integrable but have no primitive. $\endgroup$ – Ryan G Feb 14 at 15:16
  • $\begingroup$ How do we prove $A(b)=F(b)-F(a)$ under valid assumptions $\endgroup$ – Aritra Barua Feb 14 at 15:23
  • $\begingroup$ Check out these, previous discussions, and my favourite. $\endgroup$ – Ryan G Feb 14 at 15:53

They have different assumptions.

In the first part you mentioned, $f$ is assumed to be continuous. In the second part, $f$ can be assumed only Riemann integrable on the closed interval $[a,b]$. When $f$ is continuous, the second part indeed follows from the first part.

See also a comparison of the statements in this article.

  • $\begingroup$ I didn't understand,please explain to me if A(b)=F(b)-F(a) $\endgroup$ – Aritra Barua Feb 14 at 15:11
  • $\begingroup$ @AritraBarua Which part of the answer you don't understand? Please explain. $\endgroup$ – user9464 Feb 14 at 16:17
  • $\begingroup$ @AritraBarua: if you use $A(b)=F(b)-F(a)$, then you are using the first part, where $f$ is assumed to be continuous. $\endgroup$ – user9464 Feb 14 at 16:28

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