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This was an old two part exam question that I was looking over. Essentially using the improper integral $\displaystyle\int_1^9 \frac{1}{\sqrt[3]{x-1}}dx$ you are supposed to determine if the integral $\displaystyle\int_1^9 \frac{\sin^2(x)}{\sqrt[3]{x-1}} dx $ converges or diverges. If it converges find its value. I determined that $\displaystyle\lim_{t\to 1^+}\int_t^9 \frac{1}{\sqrt[3]{x-1}}dx=6$. Now I'm not sure which convergence test to use to prove that the integral diverges. Any help would be much appreciated.

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  • $\begingroup$ Comparison. Because $0 \leq \sin^2(x) \leq 1$. The integral with $0$ in the numerator is trivial. The integral with $1$ in the numerator is the one you did. So you know whether the sine integral converges or diverges and also (if it converges) an interval containing the value of its integral. $\endgroup$ Commented Jan 14, 2022 at 5:29
  • $\begingroup$ You write "$\lim_{t \rightarrow 1^+}$", but there is no "$t$" in the subsequent expression. $\endgroup$ Commented Jan 14, 2022 at 5:30
  • $\begingroup$ Now you have $\lim\limits_{x\to 1^+}$, but that too is wrong. Why? $\endgroup$ Commented Jan 14, 2022 at 6:22

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Here is a graph of the two functions.

Do you see how to proceed?

Graph of the two function

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  • $\begingroup$ From the graph it seems that both functions diverge at x=1. Is it that both functions are actually divergent? I'd also like to add that it has been about two years since I went to University. Thank you $\endgroup$ Commented Jan 14, 2022 at 14:13
  • $\begingroup$ The graph of the second function is squeezed between the graphs of the first function and the graph of $x=1$ and lies between the graphs of the first function and the graph of $y=0$, so I don't know what you mean by 'divergent' in this case. $\endgroup$ Commented Jan 14, 2022 at 16:16

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