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anonymous
 4 years ago
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.)
anonymous
 4 years ago
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.)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I already verified that the function is odd. I just need help showing that: \[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using the fact that the function is odd I know that\[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1 = 0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, I just need to show that \[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it like this \[\sin ^{2}(\sqrt{x}) ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's actually\[\sin (\sqrt[3]{x})\] or \[\sin(x^{1/3})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what did you get on this one? \[\int\limits_{}^{}\sin \sqrt{x}dx= ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[= \cos \sqrt{x} + C\]Is that correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, you need to consider (x)^1/2=u,.....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[= \cos \sqrt{x}+C\] Wouldn't that be it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we let u=x^1/2 du=1/(2x^1/2)dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0youll come up to = 2 sin x^1/2 2 x^1/2 cos x^1/2 ] from 2 to 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you have a calculator that does derivative and integrals? that way you can re check your work... :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep! I got my handydandy TI84 Plus. ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok good. did you have the same answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok good. did you have the same answer from your calculator from 2 to 3 ??
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