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Use the fact that the function is odd to show that the inequality is true. (Anyone who helps me will receive a medal because I'm stuck.)

Calculus1
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I already verified that the function is odd. I just need help showing that: \[0 \le \int\limits_{-2}^{3}\sin \sqrt[3]{x}dx \le 1\]
Using the fact that the function is odd I know that\[0 \le \int\limits_{-2}^{3}\sin \sqrt[3]{x}dx \le 1 = 0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]
Now, I just need to show that \[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]

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Other answers:

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?
Oh, okay. To begin with, it's from -2 to 3, but the integral from -2 to 0 and the integral from 0 to 2 cancel each other out this the function is odd and symmetric about the origin, so in its simplest form, the integral is from 2 to 3.
do you have a calculator that does derivative and integrals? that way you can re check your work... :D
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!

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