# Thread: average value of function

1. ## average value of function

Find the average value of the function on the interval?

H(x)= (cosx)^4sinx [0,pie]

2. Originally Posted by twilightstr
Find the average value of the function on the interval?

H(x)= (cosx)^4sinx [0,pie]
Average value of the function f(x) over the interval [a, b] is given by $\frac{\int_a^b f(x) \, dx}{b - a}$.

For your question the integral is calculated by making the substitution $u = \cos x$.

3. Originally Posted by twilightstr
Find the average value of the function on the interval?

H(x)= (cosx)^4sinx [0,pie]
Alternate method of doing the integral (from that of MrF's) is to observe that

$[\cos(x)]^4 \sin(x)=-\frac{1}{5}\frac{d}{dx} [\cos(x)]^5$

and then using the fundamental theorem of calculus.

CB

4. at the last step, i got -(cos1)^5/5pie + cos(-1)^5/5pie...
the answer in the book is 2/5pie. how was that was derived?

5. Originally Posted by twilightstr
at the last step, i got -(cos1)^5/5pie + cos(-1)^5/5pie...
the answer in the book is 2/5pie. how was that was derived?
If you post all your working it will be easier to point out the mistake(s) you have have made.

6. favg= -1/pie integral from 1 to -1 u^4du

=-1/pie[u^5/5] from 1 to -1
=-1/pie[((cos(1))^5)/5 - ((cos(-1))^5)/5]
= -((cos(1))^5)/5pie + ((cos(-1)^5)/5pie

7. Originally Posted by twilightstr
favg= -1/pie integral from 1 to -1 u^4du

=-1/pie[u^5/5] from 1 to -1

Mr F says: What follows is wrong:

=-1/pie[((cos(1))^5)/5 - ((cos(-1))^5)/5] Mr F says: It should be -1/pi[((cos(0))^5)/5 - ((cos(pi))^5)/5]. But why do this when all you have to do is evaluate the above integral??

= -((cos(1))^5)/5pie + ((cos(-1)^5)/5pie
$-\frac{1}{\pi} \int_1^{-1} u^4 \, du = \frac{1}{\pi} \int^1_{-1} u^4 \, du = \frac{1}{\pi} \left[ \frac{1}{5} u^5\right]_{-1}^1 = \frac{2}{5 \pi}$.

8. thanks, Mr Fantastic, for pointing out the flaws.