In this question we fix a real number \(\alpha\) which will be the same throughout. We say that a function \(f\) is bilateral if \[f(x) = f(2\alpha - x)\] for all \(x\).
Show that if \(f(x) = (x-\alpha)^2\) for all \(x\) then the function \(f\) is bilateral.
On the other hand, if \(f(x) = x-\alpha\) for all \(x\) then the function \(f\) is not bilateral.
Show that if \(n\) is a non-negative integer and \(a\) and \(b\) are any real numbers then \[\int_a^b x^n \:dx = -\int_b^a x^n \:dx.\]
Hence show that if \(f\) is a polynomial (and \(a\) and \(b\) are any reals) then \[\int_a^b f(x) \:dx = -\int_b^a f(x) \:dx.\]
Suppose that \(f\) is any bilateral function. By considering the area under the curve \(y = f(x)\) explain why for any \(t \geq \alpha\) we have \[\int_\alpha^t f(x) \:dx = \int_{2\alpha-t}^\alpha f(x) \:dx.\]
If \(f\) is a function we write \(G\) for the function defined by \[G(t) = \int_\alpha^t f(x) \:dx\] for all \(t\).
Suppose now that \(f\) is any bilateral function. Show that \[G(t) = -G(2\alpha-t)\] for all \(t\).
Suppose that \(f\) is a bilateral polynomial such that \(G\) is also bilateral. Show that \(G(x) = 0\) for all \(x\).
