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Determine whether the integral converges or diverges. Find the value of the integral if it converges.

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@Reaper534 can u help?
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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Other answers:

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

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