MHB Stuart's question at Yahoo Answers regarding average value of a function

[MarkFL]

Here is the question:

Find the average value of the function f on the interval [0,π]. (Enter your answer in an exact form.)?

Find the average value of the function f on the interval [0,π]. (Enter your answer in an exact form.)

f(x)=(cos^10 (x))(sin (x))

Here is a link to the question:

Find the average value of the function f on the interval [0,π]. (Enter your answer in an exact form.)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Stuart,

We are given the following theorem:

Let $f$ be continuous on $[a,b]$. The average value of $f$ on the interval is the number:

$\displaystyle A=\frac{1}{b-a}\int_a^b f(x)\,dx$

So, we identify that for our problem, we have:

$a=0,\,b=\pi,\,f(x)=\cos^{10}(x)\sin(x)$

and so we have:

$\displaystyle A=\frac{1}{\pi-0}\int_0^{\pi} \cos^{10}(x)\sin(x)\,dx$

Now, to evaluate the definite integral, we may choose the $u$-substitution:

$u=\cos(x)\,\therefore\,du=-\sin(x)\,dx$

and we now have:

$\displaystyle A=\frac{2}{\pi}\int_{0}^{1} u^{10}\,du$

Notes:

• I changed the limits in accordance with the substitution
• I used the rule $\displaystyle -\int_a^b f(x)\,dx=\int_b^a f(x)\,dx$
• I used the even function rule, i.e, given an even integrand we may write $\displaystyle \int_{-a}^a f(x)\,dx=2\int_0^a f(x)\,dx$.

Now, we may find the average value:

$\displaystyle A=\frac{2}{11\pi}\left[u^{11} \right]_0^1=\frac{2}{11\pi}\left(1^{11}-0^{11} \right)=\frac{2}{11\pi}$