4 ms·

I think the following is poorly worded and possibly incorrect: >What is the area enclosed by the function f(x) = sin²x - x²(sin³x)(cos²x) and the x-axis — betw

I think the following is poorly worded and possibly incorrect: >What is the area enclosed by the function f(x) = sin²x - x²(sin³x)(cos²x) and the x-axis — between -π/4 and π/4? Well, the second part is an odd function so we know its areas are 0 over symmetric regions, assuming "area" means "signed area". So we're left with sin²x which has a trick: the integral of sin²x and cos²x are equal over any of the quarter cycles of the graph -- same shape and everything. So we have that over our domain, pi/2 = Int(1) = Int(cos²x+sin²x) = Int(sin²x+sin²x) = 2Int(sin²x)... so the answer would be pi/4 (which doesn't appear). I also suspect that some of the answers must be wrong, possibly in the counting or probability questions where it is extremely easy to make a logical error. In my opinion the author should have a third party scrutinize his/her answers. I'm quite certain I didn't get 40% of the questions wrong. (I have a math PhD, taught calculus for 4 years, scored 170 on the quant section of the GRE, and went through this whole "test" very carefully).

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 ago
  The 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 ago
    I 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 ago
      Regardless -- 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.

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 ago
The 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 ago
  I 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 ago
    Regardless -- 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.