Logical question about the fundamental theorem of calculusSo 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 =&gt; F 1 = F 0 + C let I 1 = [ a , x ] then since; F 0 = ∫ I 1 f.We have F 0 ( a ) = 0 =&gt; F 1 ( a ) = CThen 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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Logical question about the fundamental theorem of calculusSo 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     =&amp;gt;         F    1    =      F    0    +  C let       I    1    =  [  a  ,  x  ] then since;       F    0    =      ∫                  I        1              f.We have       F    0    (  a  )  =  0  =&amp;gt;      F    1    (  a  )  =  CThen the proof is almost completed. But my problem is, if for any particular antiderivative       F    0    (  a  )  =  0 wouldn&#039;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

verzaadtwr · Accepted Answer

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 · Accepted Answer

Step 1If 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 2Noting 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  tfor any antiderivative F(x) of f(x) on [a,b].

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

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