Sarai Davenport

2022-06-28

What is the difference between the two parts of FTOC?

g2joey15

Beginner2022-06-29Added 21 answers

No, they are not the same. FTC1 is the big gun: It states that if $f$ is continuous on $[a,b],$, then f has an antiderivative $F,$, namely the function

$F(x)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in [a,b].$Recall that before the student has even seen FTC1, a lot of work has already been done in guaranteeing that the integrals${\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$ even exist (limits of Riemann sums and all that). FTC1 is a crowning acheivement in that it says not only do those integrals exist, the derivative of the function so formed gives us back $f.$.

FTC2 is a lesser acheivement. All it says is that if you have any antiderivative $G$ of a continuous function $f$ on $[a,b],$, then ${\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=G(b)-G(a).$. In FTC1 we already had an antiderivative, namely the $F$ defined there, which does this. FTC2 simply says any antiderivative will do this. The proof of FTC2 is almost trivial: By the MVT, any two antiderivatives on an interval differ by a constant, and the result follows.

$F(x)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in [a,b].$Recall that before the student has even seen FTC1, a lot of work has already been done in guaranteeing that the integrals${\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$ even exist (limits of Riemann sums and all that). FTC1 is a crowning acheivement in that it says not only do those integrals exist, the derivative of the function so formed gives us back $f.$.

FTC2 is a lesser acheivement. All it says is that if you have any antiderivative $G$ of a continuous function $f$ on $[a,b],$, then ${\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt=G(b)-G(a).$. In FTC1 we already had an antiderivative, namely the $F$ defined there, which does this. FTC2 simply says any antiderivative will do this. The proof of FTC2 is almost trivial: By the MVT, any two antiderivatives on an interval differ by a constant, and the result follows.

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