diket f'(x)=(x-1)^3 melalui titik (3,0) maka persamaan kurva tsb

f'(x) = (x - 1)³
jabarin ajah
f'(x) = x³ - 3(x²)(1) + 3(x)(1²) - 1³
f'(x) = x³ - 3x² + 3x - 1
baru integralkan
F(x) = ¼x⁴ - x³ + (3/2)x² - x + C
melalui (3,0)
F(3) = 0
F(3) = ¼(3⁴) - 3³ + (3/2)(3²) - 3/2 + C
0 = ¼(81) - 27 + 27/2 - 3/2 + C
0 = 81/4 - 108/4 + 54/4 - 6/4 + C
0 = 23/4 + C
C = -23/4
persamaan kurva nya
F(x) = ¼x⁴ - x³ + (3/2)x - x - 23/4