Logical question about the fundamental theorem of calculus So if f is continuous on [a,b] and F is

tripes3h 2022-07-08 Answered
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 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   =>   F 1 = F 0 + C let I 1 = [ a , x ] then since; F 0 = 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 = I 1 f
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Answers (2)

verzaadtwr
Answered 2022-07-09 Author has 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.
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gaiaecologicaq2
Answered 2022-07-10 Author has 6 answers
Step 1
If F(x) is any antiderivative of f(x) on [a,b], then the two functions F(x) and a x f ( t ) d t differ by at most a constant on that interval. Therefore, we have F ( x ) = a x f ( t ) d t + C for x [ a , b ].
Step 2
Noting that F ( a ) = 0 + C we have F ( x ) F ( a ) = a x f ( t ) d t, which for x = b yields the coveted equality
F ( b ) F ( a ) = a b f ( t ) d t
for any antiderivative F(x) of f(x) on [a,b].
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