I thought we cannot assume the area is symmetric because of such a graph, in which the wavy curve exists but the area covered is different. I have seen the solution, but not sure why the curve can’t be drawn like this. Please help

It’s not an even or odd function yes, but the graph is symmetric around x = 1. You can formally prove it by showing that for 0 \leq k \leq 1, f(1 + k) = f(1 - k) where f(x) = x(x - 1)(x - 2).