I read this quote attributed to VI Arnold.

“Who can’t calculate the average value of the one hundredth power of the sine function within five minutes, doesn’t understand mathematics – even if he studied supermanifolds, non-standard calculus or embedding theorems.”

EDITSource is “A mathematical trivium” A book of 100 problems that university students “should be able to solve”. The statement asks for calculation within 10% accuracy.

So the average value over the entire domain should be the same as the average value over [0,π/2]

⟨sin100(x)⟩=∫π/20sin100(x)dx∫π/20dx.

So here’s what I did:

First, this graph would be a train of highly sharp peaks. The integrand would assume values close to zero a up till before it sharply rises to 1.

So up till some ϵ∈[0,π/2] we will have sinx≈x and for the remaining π/2−ϵ interval I could find the area of triangle with base π/2−ϵ and height 1

⟨sin100(x)⟩≈2π(∫ϵ0x100dx+.5(π2−ϵ)).

I believe in principal it should be possible to find an ϵ such that the above expression yields the exact answer. So I try to approximate it, no good. Then I try mathematica and it is looking like there is no ϵ for which the value I am expecting is even close to the actual value. I plot the original and find that my approximation is hopeless.

Not to mention that my 5 minutes were over. So I admit I do not understand mathematics and humbly ask if someone could:

- Point out my mistake (Other than that ϵ is probably incomputable within 5 mins)
- How the hell is this done in 5 minutes?
The picture below has the sin100x in blue (bottom) and my approximation of it plotted against ϵ (pink). Although there is no reason for them to be together, the upper graph has a minima quite above the exact value of the integral.

EDIT

Just realizedLet u=cosx.

∫π/20sin100(x)dx=∫10(1−u2)99/2du≈∫10(1−992u2)du

**Answer**

Since sinx=eix−e−ix2i

and, for k∈Z, ∫2π0eikxdx={0k≠02πk=0,

we have ∫2π0sin100xdx=12100100∑k=0(100k)∫2π0eikx(−1)100−ke−i(100−k)xdx=(10050)21002π, and the average value is (10050)2100.

**Attribution***Source : Link , Question Author : Community , Answer Author : Andrés E. Caicedo*