tolong dong hehe pake cara yaa, thankyou!

misal luas 2 daerah itu A,
kurva y = x^2 puncak terletak pada (0,0)
luas I, 0 sampai k, luas Ii, k sampai 4
luas I = integral x^2 dx, dengan batas 0 sampai k,
= (1/3)x^3
= (1/3.)k^3 -(1/3)0^3
= (1/3)k^3
Luas Ii, integral x^2 dx. dengan batas k sampai 4.
= (1/3)x^3
= (1/3)(4)^3 - (1/3)k^3
= 64/3 - (1/3)k^3
(1/3)k^3 = 64/3 - (1/3)k^3
(1/3)k^3 + (1/3)k^3 = 64/3
(2/3)k^3 = 64/3...( kalikan 3/2)
k^3 = 32
k = (32)^1/3
k = (2^5)^1/3
k = 2^5/3
k = 2.2^2/3
k = 2^5/3