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The second part of the Fundamental Theorem of Calculus essentially states that if $$F(x) = \int^x_a{f(t)}\,dt\,,$$ then $$F'(x) = f(x)\,.$$ My question is: why does the result not depend on the lower limit of integration $a$?

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  • $\begingroup$ Novice guess here, but if $a$ is a constant, is $f(a)$ just absorbed into the constant of integration? And regardless of what it was, it would disappear upon differentiation? $\endgroup$ Jan 1, 2015 at 6:07
  • $\begingroup$ You already defined F(x) in terms of the integral with its lower bound anyways $\endgroup$
    – Quality
    Jan 1, 2015 at 6:07
  • $\begingroup$ An excellent answer has been provided. $\endgroup$
    – MPW
    Jan 1, 2015 at 6:18
  • $\begingroup$ a is a starting point and is a fixed constant ... flip the question around and ask yourself, why should it matter? $\endgroup$
    – oemb1905
    Oct 12, 2017 at 21:38

1 Answer 1

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Let, e.g., $b<a$. Then $$ G(x) := \int^x_b{f(t)}\,dt=\int_b^a {f(t)}\,dt+\int_a^x {f(t)}\,dt=F(x)+D, $$ and the derivatives of $F$ and $G$ are identical.

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    $\begingroup$ Nice, succinct. +1. $\endgroup$
    – MPW
    Jan 1, 2015 at 6:17

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