tripes3h

2022-07-08

Logical question about the fundamental theorem of calculus
So if f is continuous on [a,b] and F is any antiderivative of f, and $I=\left[a,b\right]$ then ${\int }_{I}f=F\left(b\right)-F\left(a\right)$.
Okay So the main proof for this as follows :
Let ${F}_{0},{F}_{1}$ be particular antiderivatives, We have let ${I}_{1}=\left[a,x\right]$ then since; ${F}_{0}={\int }_{{I}_{1}}f$.
We have ${F}_{0}\left(a\right)=0=>{F}_{1}\left(a\right)=C$
Then the proof is almost completed. But my problem is, if for any particular antiderivative ${F}_{0}\left(a\right)=0$ wouldn't ${F}_{1}\left(a\right)=0$ be the case, since what we did here was to use the definition of antiderivative, we could easily write ${F}_{1}={\int }_{{I}_{1}}f$

Explanation:
That does not happen to any particular derivative. You did not pick ${F}_{0}$ to be any antiderivative, you picked the one that evaluates the integral of f starting from the point a, that is unique. In other words, you did not only used the definition of antiderivative, as you said, you picked a specific one.

gaiaecologicaq2

Step 1
If F(x) is any antiderivative of f(x) on [a,b], then the two functions F(x) and ${\int }_{a}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$ differ by at most a constant on that interval. Therefore, we have $F\left(x\right)={\int }_{a}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt+C$ for $x\in \left[a,b\right]$.
Step 2
Noting that $F\left(a\right)=0+C$ we have $F\left(x\right)-F\left(a\right)={\int }_{a}^{x}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$, which for $x=b$ yields the coveted equality
$F\left(b\right)-F\left(a\right)={\int }_{a}^{b}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt$
for any antiderivative F(x) of f(x) on [a,b].

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