3 ms·
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 inte
by SamReidHughes 6y ago
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].
- virtuous_signal 6y agoThe second summand due to the sin^3 is not nonnegative (so you do need to use oddness to avoid the integral) Right on the second part, my bad.
- SamReidHughes 6y agoI read that part of your comment as re-interpreting the question, instead of just reinterpreting the prior part of the sentence it's in, never mind.
- virtuous_signal 6y agoRegardless -- I think it's a valid reading to say the "area between" say, g(x) = x and the x-axis from -1 to 1 equals 1. Only people who are trained would say it's 0. So I don't think something like that really measures "extended problem solving and deductive thinking skills" which is a pretty lofty goal for a test.