# Simple proof needed

1. May 3, 2007

### John O' Meara

Prove that $$\int_0^a f(x)dx = \int_0^{\frac{a}{2}}[f(x) + f(a-x)]dx \\$$
Hence prove that $$\int_0^{\pi} x sin^6x dx = \pi \int_0^{\frac{\pi}{2}} sin^6x dx \\$$ and evaluate this integral using the following reduction formula $$I_{m,n}= \frac{n-1}{m+n}I_{m,n-2} \\$$

My effort:
$$\int_0^a f(x)dx = \int_0^{\frac{a}{2}}f(x)dx + \int_{\frac{a}{2}} f(a-x)dx \\$$. Now
$$\int_{\frac{a}{2}}^a f(x)dx = \int_{\frac{a}{2}}^a f(a-x)dx \\$$ ... by a theorem. Therefore $$\int_0^a f(x)dx = \int_0^{\frac{a}{2}} f(x)dx + \int_{\frac{a}{2}}^a f(a-x)dx \\$$. I need help to change the integral's limits to 0 and a/2 on the second integral on the right, thanks very much.

2. May 3, 2007

### MathematicalPhysicist

$$\int_{0}^{a}f(x)dx$$
change variable u=a-x then we have du=-dx
$$-\int_{a}^{0}f(u)du=-\int_{0}^{a}f(a-x)dx$$
so now you have: $$\int_{0}^{a}f(x)dx=\int_{0}^{a}[f(x)-f(a-x)]/2dx$$
another change u=x-(a/2), which gives you:
$$\int_{-a/2}^{a/2}f(u+a/2)-f(a/2-u)]/2du$$
this function in the integrand is an odd function, which means that this equals 2 times the integral but from 0 to a/2.
if im mistaken, sorry for misleading.

Last edited: May 3, 2007
3. May 4, 2007

### Gib Z

You are mistaken :( $$\int^a_0 f(x) dx = \int^a_0 f(a-x) dx$$

On your 2nd line of latex, Reversing the bounds as you did makes the integrals value negative, canceling the other negative.

Jean - To finish it off, u=x - (a/2) for the last integral.

4. May 4, 2007

### MathematicalPhysicist

well it should be $$-\int_{a}^{0}f(a-u)du=\int_{0}^{a}f(a-u)du$$
so yes, you are right, thanks for the correction.