Question

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.)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

Ah, u substitution.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

is it from -2 to 3?

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

ok good. did you have the same answer?

Answer 18

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

Answer 19

Yep. Thank you!