graph{4 - x^2 [-0.5, 4, -8, 5.315]}
This requires equating two integrals, one for the area from #x =0# to #x = 2#, and another for the area from #x = 2# and #x = p#, for some #p#. (Note: As noted in the graph above, the "area" between 2 and #p# will be negative. I'm not certain about the note in your question regarding that area being considered positive, as that isn't the typical way this problem is approached. You can handle this by either negating the integral from 2 to #p#, or by calculating the integral "in reverse" from #p# to #2#:
#A = int_0^2(4-x^2) dx = (4x-x^3/3)|_0^2=8-8/3=16/3#
#B = int_p^2(4-x^2) dx = (4x-x^3/3)|_p^2#
#=(8-8/3)-(4p-p^3/3)=16/3-4p+p^3/3#
Per the problem, we seek #p# such that #A = B#:
#16/3 = 16/3-4p+p^3/3#
#0 = -4p+p^3/3#
#0 = p^3/3-4p#
#0 = p(p^2/3-4) => p=0 or p^2/3-4=0#
We ignore the result #p=0# as we were told that #p# lies beyond #x = 2#. Thus:
#p^2/3-4=0#
#p^2-12=0#
#p^2=12 => p=+-sqrt(12) = +-2sqrt(3)#
For similar reasons to why we rejected #p = 0#, we can reject #p = -2sqrt(3)#.
The answer is #p = 2sqrt(3)#.
