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Now, I just need to show that \[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]

is it like this
\[\sin ^{2}(\sqrt{x}) ?\]

It's actually\[\sin (\sqrt[3]{x})\] or \[\sin(x^{1/3})\]

what did you get on this one?
\[\int\limits_{}^{}\sin \sqrt{x}dx= --------------?\]

\[= \cos \sqrt{x} + C\]Is that correct?

no, you need to consider (x)^1/2=u,.....

Ah, u substitution.

\[= -\cos \sqrt{x}+C\] Wouldn't that be it?

if we let u=x^1/2
du=1/(2x^1/2)dx

youll come up to
= 2 sin x^1/2 -2 x^1/2 cos x^1/2 ] from 2 to 3

is it from -2 to 3?

Yep! I got my handy-dandy TI-84 Plus. ;)

ok good. did you have the same answer?

ok good. did you have the same answer from your calculator from 2 to 3 ??

Yep. Thank you!