Question

Find the average value of f(x)=7sinx+8cosx on the interval [0,19pi/6]

Answer 1

So I believe, first you would take the integral: \[\int\limits_{0}^{19\pi/6}1 dx\] which comes out to be 19pi/6.

Then evaluate \[\int\limits_{0}^{19\pi/6} 7sinx+8cosx dx\] You would then divide the value of the second integral by the value of the first integral. Do you have a way to verify if this answer's correct?

Answer 2

I keep getting the wrong answer?

Answer 3

(1/(b-a))*integral of f(x) from a to b

Answer 4

\[\int\limits_{0}^{19\pi/6}7sinx+8cosxdx\]= [-7cosx+8sinx] from 0 to \[19\pi/6\] Then plug in the values. Lastly, divide by \[19\pi/6\] My final answer is something like 9.06.

Answer 5

yeah, thats what I got but its wrong.

Answer 6

ave = (1/(19*pi/6))*(-7cos(19*pi/6)+8sin(19*pi/6))
=[3*(7*sqrt(3)+6)]/19*pi
=.91092 i think

Answer 7

(1/(19*pi/6))*[(-7cos(19*pi/6)+8sin(19*pi/6)-(-7cos(0)]*

Answer 8

Are you using WebAssign, by any chance? Alternatively, the answer can be expressed as: \[3+\frac{ \sqrt{7} }{ 2 }\]