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State whether the following integral is convergent or divergent:$$\int_{0}^{1} \frac{\sin(x)}{x^{1.5}} \ dx $$

The answer says that it converges due to a comparison with $\frac{1}{\sqrt{x}}$. I don't see how this works, as $\frac{1}{\sqrt{x}}$ is < $\frac{\sin(x)}{x^{1.5}}$ for $0 < x < 1$.

Any help will be greatly appreciated, thanks in advance.

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Note that for $x\in [0,1]$, $0\le \sin(x)\le x $. Hence, we have

$$0\le \frac{\sin(x)}{x^{3/2}}\le \frac{1}{\sqrt x}$$

Can you finish now?

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  • $\begingroup$ Yes. Thank you! $\endgroup$ – StopReadingThisUsername Nov 16 '17 at 4:12
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Nov 16 '17 at 14:21

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