Question

Average value of a function on an interval.

Will post equstion and work in a comment

Answer 1

\[\frac{ 1 }{ \pi - 0 } \int\limits\limits_{0}^{\pi} \sin (x) dx\]
\[\int\limits_{0}^{\pi} \sin(x) dx = [-\cos] _{0}^{\pi}\]
\[[-\cos \pi] - [-\cos 0] => [-(-1)] - [-1] = 1 + 1 = 2\]
Average value =
\[\frac{ 1 }{ \pi } (2) = \frac{ 2 }{ \pi }\]
I'm asked to find the x value that corresponds to 2/pi

sin(x) = 2/pi

using the arcsin on the calculator I get ~ 0.690

But there is a second x value that corresponds to 2/pi, which is 2.451, how do I solve for the 2nd x value?

Answer 2

Can I find the 2nd x value algebraically, using calculus, or using a graphing calculator?

Answer 3

@rushwer

Answer 4

@Rushwr

Answer 5

it's going to be symmetrical about \(\pi/2\)