THEOREM 6 If $f$ is integrable on $[a,b]$, then for any number $c$, the function $cf$ is integrable on $[a,b]$ and $$\int_a^b cf=c\cdot\int_a^b f.$$

PROOF The proof (which is much easier than that of Theorem 5) is left to you. It is a good idea to treat separately the cases $c\geq 0$ and $c\leq 0$. Why?

I wonder why the author Michael Spivak wrote "Why?".
Is there any deep reason?

My Proof :
I think it is very natural to treat separately the cases $c=0$ and $c>0$ and $c<0$ because

  • it is obvious that $\int_a^b cf=0=c\cdot\int_a^b f$ if $c=0$ and
  • $L(cf,P)=cL(f,P)$ and $U(cf,P)=cU(f,P)$ holds if $c>0$ and
  • $L(cf,P)=cU(f,P)$ and $U(cf,P)=cL(f,P)$ holds if $c<0$.
  • 3
    $\begingroup$ sounds fine to me $\endgroup$
    – Andrew
    Nov 27, 2023 at 4:19
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    $\begingroup$ Ditto. ${}{}{}$ $\endgroup$
    – copper.hat
    Nov 27, 2023 at 4:19
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    $\begingroup$ Do you really need to handle the case $c=0$ separately though? If we can trim the proof slightly with zero extra effort we might as well do so. $\endgroup$
    – littleO
    Nov 27, 2023 at 4:19
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    $\begingroup$ Andrew, copper.hat and littleO, thank you very much for your comments. $\endgroup$ Nov 27, 2023 at 4:34
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    $\begingroup$ I think the Question is not whether we want to club $0$ with the other cases or make it a new case. It is about why Spivak is including the "why" when suggesting to make it separate cases. In other words , "why not Prove it for all $c$ in one shot?" : +1 for valid Query. It is okay to make it 2 cases or even 3 cases. Why Spivak is suggesting to not make it 1 Case ? OP has the Proof , which looks right. $\endgroup$
    – Prem
    Nov 27, 2023 at 6:52

1 Answer 1


This is a Soft Question & the Author is unfortunately not around to give the Definitive Answer.

The way I am looking at it , it is because of the way the integral was given on Page 258 & Page 259 :

Page 258

Page 259

It is using $\le$ everywhere.
When we want to Prove theorem 6 (or whichever exercise) , we have to be careful about that.
Of course , we can multiply both sides by Positive (or even $0$) $c$ while keeping that relation unchanged.
When multiplying both sides by Negative (or even $0$) $c$ , we have to change the relation to $\ge$ or we have to switch the sides.

Tracking that is easier when we have separate cases.
Thus Spivak is suggesting that.
It is a Suggestion & we can work without that , though it might be slightly harder to keep track at early learning stages where that complication is unnecessary & avoidable.

Proving the earlier theorems [ including theorem 5 & theorem 2 ] necessitates using $\le$ & $\ge$ : Spivak is very careful when using those.

Proof can alternately consider Positive $c$ & $c=0$ & then $c=-1$ : That too will cover all Cases , though it is more work.


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