# Fundamental Theorem of Calculus: Why Doesn't the Integral Depend on Lower Bound?

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 $a$?

• 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? – turkeyhundt Jan 1 '15 at 6:07
• You already defined F(x) in terms of the integral with its lower bound anyways – Quality Jan 1 '15 at 6:07
• An excellent answer has been provided. – MPW Jan 1 '15 at 6:18
• a is a starting point and is a fixed constant ... flip the question around and ask yourself, why should it matter? – oemb1905 Oct 12 '17 at 21:38

## 1 Answer

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.

• Nice, succinct. +1. – MPW Jan 1 '15 at 6:17