Calculus problem

[Steven G]

You can easily see that [math]\int_0^{2\pi}sinx dx=0[/math]
Can anyone see why [math]\int_0^{2\pi}xcosx dx=0?[/math]
An algebraic argument will be just fine.

 

[Harry_the_cat]

Do you mean WITHOUT using integration by parts?

 

[blamocur]

Spoiler

If we can easily see that [imath]\int_0^{2\pi} \cos x dx = 0[/imath] then [imath]\int_0^{2\pi} x\cos x dx = \int_0^{2\pi} (x-\pi) \cos x dx[/imath], but the latter is anti-symmetric relative to [imath]x=\pi[/imath].

 

[Steven G]

Do you mean WITHOUT using integration by parts?

Yes

 

[Steven G]

Spoiler

If we can easily see that [imath]\int_0^{2\pi} \cos x dx = 0[/imath] then [imath]\int_0^{2\pi} x\cos x dx = \int_0^{2\pi} (x-\pi) \cos x dx[/imath], but the latter is anti-symmetric relative to [imath]x=\pi[/imath].

[imath]\int_0^{2\pi} x\cos x dx = \int_0^{2\pi} (x-\pi) \cos x dx =\int_0^{2\pi} x \cos x dx-\pi\int_0^{2\pi} x\cos x dx[/imath] where this last integral is zero. Is this what you meant? If yes, it doesn't seem to help.

 

[blamocur]

[imath]\int_0^{2\pi} x\cos x dx = \int_0^{2\pi} (x-\pi) \cos x dx =\int_0^{2\pi} x \cos x dx-\pi\int_0^{2\pi} x\cos x dx[/imath] where this last integral is zero. Is this what you meant? If yes, it doesn't seem to help.

Spoiler

What I meant is that [imath]f(x) = (x-\pi)\cos x[/imath] is anti-symmetric relative to [imath]\pi[/imath], i.e. [imath]f(x) = -f(2\pi-x)[/imath]. This means that in the partial sums (used in the definition of definite integrals) each element is canceled by the symmetrical element and all partial sums are 0.
This can also be achieved by breaking the integral into two: [imath]\int_0^{2\pi} f(x)dx = \int_0^{\pi} f(x)dx + \int_\pi^{2\pi} f(x)dx[/imath], and then replacing the variable in the second integral: [imath]u=2\pi-x[/imath]