diketahui f (x) = 2x-10, f' (2) = -18 dan f(0)= 5. nilai f(3) adalah

f "(x) = 2x - 10,

maka
f '(x)
= ∫ 2x - 10  dx
= x² - 10x + c

f '(2) = -18
2² - 10(2) + c = -18
4 - 20 + c = -18
c = -18 + 16
c = -2

f(x)
= ∫ f '(x) dx
= ∫ x² - 10x - 2  dx
= x³/3 - 5x² - 2x + C

f(0) = 5
C = 5

jadi 
f(3)
= 3³/3 - 5(3)² - 2(3) + 5
= 9 - 45 - 6 + 5
= -37