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Please how do find $b\gt 1$ so that $$ \int_1^b (x-2)^3~dx =0?$$ This question is on a chapter dealing with antiderivatives and I'm not sure how to go about it. At this point it is assumed that I don't know how to integrate yet. I'm also not allowed to use the fundamental theorem of calculus.

vonbrand
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Gorg
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1 Answers1

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Note that the substitution $u=x-2$ gives the integral $$\int_{-1}^{b-2}u^3\,du.$$ What do you know about integrals of odd functions (in particular what sort will give an integral of $0$)?

The fact that $u\mapsto u^3$ is strictly monotone is enough to show uniqueness of the value thus determined. (Thanks to Fly by Night for pointing out that something additional was required to prove uniqueness.)

Cameron Buie
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  • if $b-2=-1$ and hence $b=3$. – Gorg Apr 04 '13 at 19:39
  • I think you mean "$b-2=1$", but that's it exactly! – Cameron Buie Apr 04 '13 at 19:40
  • right. Thanks very much. – Gorg Apr 04 '13 at 19:42
  • @Gorg Being an odd function is not sufficient. The key is that $u^3 > 0 \iff u > 0$. – Fly by Night Apr 04 '13 at 19:54
  • @FlybyNight: Given any $L>0$ and any odd function $f$ integrable on $[-L,L]$, we have $$\int_{-L}^Lf(u),du=0.$$ – Cameron Buie Apr 04 '13 at 20:06
  • @FlybyNight: In point of fact, the condition that you proffer is not sufficient. Let $$f(u)=(u+1)^3-1.$$ Then $f(u)>0$ iff $u>0,$ but given any $L>0$, $$\int_{-L}^Lf(u),du>O.$$ – Cameron Buie Apr 04 '13 at 20:24
  • @CameronBuie First of all, I didn't suggest my condition was sufficient; it was simply a demonstration that yours wasn't. I agree that if the limits are symmetrical, i.e. $-L<x<L$, then the definite integral is zero. However, the post-substitution limits are not symmetrical. Of course $b=1$ makes the limits symmetrical, but we do not know that this value of $b$ is unique. We need my condition. – Fly by Night Apr 04 '13 at 20:40
  • @FlybyNight: The point of my hint was to find $b$ such that the post-substitution limits were symmetrical. – Cameron Buie Apr 04 '13 at 20:44
  • @FlybyNight: I see what you were getting at, now. For uniqueness, we do need more. – Cameron Buie Apr 04 '13 at 22:05
  • @CameronBuie That's right. Thanks for your reply; it's appreciated. – Fly by Night Apr 05 '13 at 17:29