anonymous
  • anonymous
Determine whether the integral converges or diverges. Find the value of the integral if it converges.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
@Reaper534 can u help?
terenzreignz
  • terenzreignz
Let's just focus on the integral first, keeping in mind that... \[\Large \int\limits_1^\infty x^{-\frac43}= \lim_{b\rightarrow\infty}\int\limits_1^bx^{-\frac43}\]

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anonymous
  • anonymous
okay
terenzreignz
  • terenzreignz
Well, can you integrate \[\Large \int x ^{-\frac43}=\color{red}?\]
anonymous
  • anonymous
yes i get -3/3sqrt(x)
terenzreignz
  • terenzreignz
This \[\Large \frac{-3}{3\sqrt{x}}\]?
anonymous
  • anonymous
yes!
terenzreignz
  • terenzreignz
Or this \[\Large \frac{-3}{\sqrt[3]x}\]
anonymous
  • anonymous
the second one
terenzreignz
  • terenzreignz
lol... it's cube root, not 3sqrt because \(\Large3\sqrt x\) means something entirely different, ok?
anonymous
  • anonymous
Okay
terenzreignz
  • terenzreignz
Okay, so, we have... \[\Large \int\limits_1^b x^{-\frac43}=\left.\frac{-3}{\sqrt[3]x}\right]_1^b\]
terenzreignz
  • terenzreignz
Can you evaluate this bit?
anonymous
  • anonymous
i think
terenzreignz
  • terenzreignz
Well then, what do you get?
anonymous
  • anonymous
hmm i'm working on it
anonymous
  • anonymous
eh...i'm getting a wrong answer :/
terenzreignz
  • terenzreignz
Fundamental theorem of Calculus? \[\Large \int\limits_a^b f'(x)dx = \left.f(x)\right]_a^b = f(b)-f(a)\]
anonymous
  • anonymous
ohhhh okay
terenzreignz
  • terenzreignz
So... \[\Large \int\limits_1^b x^{-\frac43} \ dx=\left.\frac{-3}{\sqrt[3]x}\right]_1^b=\color{red}?\]
anonymous
  • anonymous
so i plug in b and 1 into that then subtract?
terenzreignz
  • terenzreignz
Yes... but that's not the end yet, just plug in for now, and tell me what you get :)
anonymous
  • anonymous
okay so -3/3sqrt(b) - -3/3sqrt(1)
terenzreignz
  • terenzreignz
I'm assuming by 3sqrt you mean cube root :D Okay, that being the case, you're right :) \[\Large \int\limits_1^b x^{-\frac43}=\frac{-3}{\sqrt[3]b}+\frac3{\sqrt[3]1}= -\frac3{\sqrt[3]b}+3\] Catch me so far?
anonymous
  • anonymous
yes
terenzreignz
  • terenzreignz
Now, we're supposed to take the improper integral to infinity, right? Remember this... \[\Large \int\limits_1^\infty x^{-\frac43} \ dx = \color{red}{\lim_{b\rightarrow\infty}}\int\limits_1^bx^{-\frac43} \ dx\] Now is the time to apply that limit (which we haven't done yet) \[\Large \color{red}{\lim_{b\rightarrow\infty}}\left(-\frac3{\sqrt[3]b}+3\right) \]
anonymous
  • anonymous
okay
terenzreignz
  • terenzreignz
So... evaluating the limit...?
anonymous
  • anonymous
umm i get 3? :/
terenzreignz
  • terenzreignz
That's right!
anonymous
  • anonymous
ohh okay lol
anonymous
  • anonymous
so thats the final answer?
terenzreignz
  • terenzreignz
Well, you technically have two questions, but since there was an answer, then the integral converges, and it converges to 3 ^.^
anonymous
  • anonymous
okay thanks
terenzreignz
  • terenzreignz
No problem :)

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