Finding Lines "a" & "b" through Integration 1. The problem statement, all variables and given/known data Let f be the function given by f(x)=4cosx a) This one I figured out. b) Find the numbers a and b such that the lines x=a and x=b divide the region R into three regions with equal area. 2. Relevant equations ∫4cos(x)dx=4sin(x) + c 3. The attempt at a solution Typing out the question, I started to realize one thing; a and b will be numbers, not variables. So, these lines where x=a and x=b will be vertical. Knowing this is much more helpful. But now that I have this logic deduced, let me try and see if I could do the problem. Hm.... Oh I know what else would help.... a very important yet easy to miss rule...The sum rule of integration. So, ∫[0,a] 4cos(x)dx + ∫[a,b] 4cos(x)dx + ∫[b,∏/2] 4cos(x)dx= 4 since from letter a question which I didn't state is 4 (∫[0,∏/2]4cos(x)dx=4) Therefore, ∫4cos(x)dx=4sin(x) + c & assuming c=0 *is this a safe assumption? I would believe so, but I can't seem to prove it 100% [4sin(a)] + [(4sin(b)-4sin(a)] + [(4sin(∏/2)-4sin(b))]=4 but also ∫[0,a] 4cos(x)dx = ∫[a,b] 4cos(x)dx = ∫[b,∏/2] 4cos(x)dx= 4/3 (three equal areas) then I could solve for b and a hopefully. Let me come up with some equations... so, 4sin(b) - 4sin(a)=4/3 4sin(b) - 4/3 = 4sin(a) ∫[0,a] 4cos(x)dx=4sin(a)=4/3 therefore, 4sin(b) - 4/3 = 4/3 4sin(b)=8/3 sin(b)=2/3 b=sininverse(2/3) b= .7297=x so, ∫[a,b] 4cos(x)dx=[(4sin(b)-4sin(a)]=4/3 4sin(.7297) - 4sin(a)=4/3 a=sininverse(.33333333)=.3398 Does this look right everyone? a=.3398, b=.7297 it looks right to me. I guess when I attempted the solution, I wasn't thinking right or something.