Question

Verify that the function is an odd function and use that fact to show that the inequality is true.

Answer 1

Verify that \[f(x)=\sin \sqrt[3]{x}\] is an odd function and use that fact to show that \[0 \le \int\limits_{-2}^{3}\sin \sqrt[3]{x}dx \le 1\]

Answer 2

odd function means f(-x) = - f(x)
\[f(-x)=\sin \sqrt[3]{-x} = \sin -\sqrt[3]{x} = -\sin \sqrt[3]{x}\]

Answer 3

Okay. I got that. Now, how do I use that fact to show that the inequality is true?

Answer 4

becomause it's odd
\[ \int\limits\limits_{-2}^{0}\sin \sqrt[3]{x}dx = \int\limits\limits_{0}^{2}\sin \sqrt[3]{x}dx\]

Answer 5

so they cancel out leaving
\[\int\limits\limits_{2}^{3}\sin \sqrt[3]{x}dx \]

Answer 6

now you just need to show
\[0\le\int\limits\limits\limits_{2}^{3}\sin \sqrt[3]{x}dx \le1\]

Answer 7

Okay. That's what I figured from looking at the graph and from what I know about integrals of odd functions that are centered about the origin. Let me try to finish it, and if I get stuck I'll post another reply.

Answer 8

I'm stuck. I'm kind of going in circles with the last part. How do I show this? \[0 \le \int\limits_{2}^{3}\sin \sqrt[3]{x}dx \le 1\]