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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asked Jan 1, 2015 at 6:01
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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? $\endgroup$– turkeyhundtCommented Jan 1, 2015 at 6:07
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$\begingroup$ You already defined F(x) in terms of the integral with its lower bound anyways $\endgroup$– QualityCommented Jan 1, 2015 at 6:07
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$\begingroup$ An excellent answer has been provided. $\endgroup$– MPWCommented Jan 1, 2015 at 6:18
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$\begingroup$ a is a starting point and is a fixed constant ... flip the question around and ask yourself, why should it matter? $\endgroup$– oemb1905Commented Oct 12, 2017 at 21:38
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1 Answer
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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.
answered Jan 1, 2015 at 6:08
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