# Why did the author write "Why?"? Any deep reason? ("Calculus Fourth Edition" by Michael Spivak)

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 $$c<0$$.
• sounds fine to me Nov 27, 2023 at 4:19
• Ditto. ${}{}{}$ Nov 27, 2023 at 4:19
• 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. Nov 27, 2023 at 4:19
• Andrew, copper.hat and littleO, thank you very much for your comments. Nov 27, 2023 at 4:34
• 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.
– Prem
Nov 27, 2023 at 6:52

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 :

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.