Finding mean values of a function

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The question is :- Find the mean value of (cos 2x)7 with respect to x over the interval 0 ≤ x ≤ 0.25(pi), leaving your answer in terms of (pi).


I just don't know the formula for calculating this so if anyone call tell me that hopefully I will be able to solve this question by myself :)
 
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rdajunior95 said:
The question is :- Find the mean value of (cos 2x)7 with respect to x over the interval 0 ≤ x ≤ 0.25(pi), leaving your answer in terms of (pi).


I just don't know the formula for calculating this so if anyone call tell me that hopefully I will be able to solve this question by myself :)

The mean value of a function f over an interval [a, b] is defined as
\frac{1}{b - a}\int_a^b f(x) dx

Since there's an integral involved, this should probably have been posted to the Calculus & Beyond section.
 
Thanks, I will try it out and see if I get the answer :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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