## nathanruff 2 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. nathanruff

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. nathanruff

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. nathanruff

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

4. mark_o.

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

5. nathanruff

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

6. mark_o.

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

7. nathanruff

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

8. mark_o.

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

9. nathanruff

Ah, u substitution.

10. nathanruff

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

11. mark_o.

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

12. mark_o.

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

13. mark_o.

is it from -2 to 3?

14. nathanruff

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. mark_o.

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

16. nathanruff

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

17. mark_o.

ok good. did you have the same answer?

18. mark_o.

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

19. nathanruff

Yep. Thank you!