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 $I=[a,b]$ then ${\int}_{I}f=F(b)-F(a)$.

Okay So the main proof for this as follows :

Let ${F}_{0},{F}_{1}$ be particular antiderivatives, We have ${F}_{1}-{F}_{0}=C\text{}=\text{}{F}_{1}={F}_{0}+C$ let ${I}_{1}=[a,x]$ then since; ${F}_{0}={\int}_{{I}_{1}}f$.

We have ${F}_{0}(a)=0=>{F}_{1}(a)=C$

Then the proof is almost completed. But my problem is, if for any particular antiderivative ${F}_{0}(a)=0$ wouldn't ${F}_{1}(a)=0$ be the case, since what we did here was to use the definition of antiderivative, we could easily write ${F}_{1}={\int}_{{I}_{1}}f$

So if f is continuous on [a,b] and F is any antiderivative of f, and $I=[a,b]$ then ${\int}_{I}f=F(b)-F(a)$.

Okay So the main proof for this as follows :

Let ${F}_{0},{F}_{1}$ be particular antiderivatives, We have ${F}_{1}-{F}_{0}=C\text{}=\text{}{F}_{1}={F}_{0}+C$ let ${I}_{1}=[a,x]$ then since; ${F}_{0}={\int}_{{I}_{1}}f$.

We have ${F}_{0}(a)=0=>{F}_{1}(a)=C$

Then the proof is almost completed. But my problem is, if for any particular antiderivative ${F}_{0}(a)=0$ wouldn't ${F}_{1}(a)=0$ be the case, since what we did here was to use the definition of antiderivative, we could easily write ${F}_{1}={\int}_{{I}_{1}}f$

verzaadtwr

Beginner2022-07-09Added 17 answers

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.

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

Beginner2022-07-10Added 6 answers

Step 1

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

Step 2

Noting that $F(a)=0+C$ we have $F(x)-F(a)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, which for $x=b$ yields the coveted equality

$F(b)-F(a)={\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$

for any antiderivative F(x) of f(x) on [a,b].

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

Step 2

Noting that $F(a)=0+C$ we have $F(x)-F(a)={\int}_{a}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, which for $x=b$ yields the coveted equality

$F(b)-F(a)={\int}_{a}^{b}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$

for any antiderivative F(x) of f(x) on [a,b].

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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