## 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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1. anonymous

I already verified that the function is odd. I just need help showing that: $0 \le \int\limits_{-2}^{3}\sin \sqrt[3]{x}dx \le 1$

2. anonymous

Using the fact that the function is odd I know that$0 \le \int\limits_{-2}^{3}\sin \sqrt[3]{x}dx \le 1 = 0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1$

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

Ah, u substitution.

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

is it from -2 to 3?

14. anonymous

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.

15. anonymous

do you have a calculator that does derivative and integrals? that way you can re check your work... :D

16. anonymous

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

17. anonymous

ok good. did you have the same answer?

18. anonymous

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

19. anonymous

Yep. Thank you!