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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.

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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.

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  • $\begingroup$ Ah yes this is true. Is there an argument for $f$ not continuous? $\endgroup$
    – Saran Wrap
    Commented Apr 10, 2020 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
    Commented Apr 10, 2020 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
    Commented Apr 10, 2020 at 18:22

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