The function is non-negative on [-pi/4, pi/4], so you don't need any assumption about area meaning "signed area."
Your math is wrong on that problem. The integral of sin^2 over [-pi/4, pi/4] is (pi - 2)/4, which was one of the options. The functions sin^2 and cos^2 only have equal integrals over properly aligned quarter-cycles like [0, pi/2].