If f(x) ≡ f(a - x), prove 2 * ∫(from 0 to a) x * f(x) dx = a * ∫(from 0 to a) f(x) dx?

Here is a proof via substitution:

Let x = a - u.

So, u = a - x, and du = -dx.

Bounds:

x = 0 ==> u = a, and x = a ==> u = 0.

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So, ∫(x = 0 to a) x f(x) dx

= ∫(u = a to 0) (a - u) f(a - u) * -du

= ∫(u = 0 to a) (a - u) f(a - u) du, via ∫(a to b) f dx = -∫(b to a) f dx

= ∫(u = 0 to a) (a - u) f(u) du, via hypothesis that f(u) = f(a - u)

= ∫(x = 0 to a) (a - x) f(x) dx, dummy variable change

= ∫(x = 0 to a) (a f(x) - x f(x)) dx

= a * ∫(x = 0 to a) f(x) - ∫(x = 0 to a) x f(x) dx, via linearity.

Hence, we have ∫(x = 0 to a) x f(x) dx = a * ∫(x = 0 to a) f(x) - ∫(x = 0 to a) x f(x) dx.

Solve for ∫(x = 0 to a) x f(x) dx by adding it to both sides:

2 * ∫(x = 0 to a) x f(x) dx = a * ∫(x = 0 to a) f(x) dx.

I hope this helps!

If the funtion is an even funtion. ie, the degree is even. The funtion mirrors itself on the y-axis. you can multiply the integral by 2 and go from 0 to a, rather than -a, a.