Spivak states the following (from page 370 to 371 in fourth edition of Calculus): "If we can find $\int_{a}^{b}f(x)dx$ for all $a$ and $b$, then we can certainly find $\int f(x)dx$." In other words knowing the indefinite integral for all possible $a$ and $b$ implies we have an anti-derivative of $f$ - a function $F$ such that $F'=f$. But how do you prove this (I'm not sure how to even show that such a $F$ is differentiable)?

The example he provides is the following: Since for all $a, b$ $$\int_{a}^{b}\sin^{5}x \cos x \, dx = \frac{\sin^{6} b}{6} - \frac{\sin^{6} a}{6}$$ then this implies $$\int \sin^{5}x \cos x \, dx = \frac{\sin^{6} x}{6}$$ The example makes sense, but why does it apply to $f$ in general?

I always thought you went from using indefinite integrals to evaluate definite integrals, so I'm confused as to why we can go backwards.


Remember that one of the basic steps in proving the Fundamental Theorem of Calculus is that if you have a continuous function $f$, then the function $G$ defined as $G(t)=\int_a^tf(x)dx$ is an antiderivative of $f$.

How was it proved? You form the difference quotient for $G'(t)$: $$ G'(t)=\lim_{h\to0}\frac{G(t+h)-G(t)}h=\lim_{h\to0}\frac{\int_a^{t+h}f(x)dx-\int_a^tf(x)dx}h =\lim_{h\to0}\frac{\int_t^{t+h}f(x)dx}h $$ and how do you look at that? If you want to think of the integral as measuring area, then you’re dividing the area of a thin strip that’s nearly $f(t)$ high and $h$ wide, by $h$.

And there you are.

  • $\begingroup$ Ah yes this is true. Is there an argument for $f$ not continuous? $\endgroup$ – Saran Wrap Apr 10 '20 at 4:00
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    $\begingroup$ @SaranWrap: if $f$ has a left limit $L$ at $c$ then left derivative of $\int_{a} ^{x} f(t) \, dt$ at $c$ equals $L$. A similar statement holds for right limit and right derivative. Thus if $f$ has a jump discontinuity at $c$ then the function $G$ in this answer is not differentiable at $c$. The really strange case is when $f$ has oscillatory kind of discontinuity at $c$. Here $G$ may or may not have a derivative and FTC can't distinguish between these. $\endgroup$ – Paramanand Singh Apr 10 '20 at 8:25
  • $\begingroup$ À propos the comment of @ParamanandSingh , there seems to be a theorem (I’ve not seen a proof, nor a formal statement) that a derivative-function, if it exists, must satisfy the intermediate-value condition, so can be discontinuous, but in a very special way. $\endgroup$ – Lubin Apr 10 '20 at 18:22

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