tolong ya bantuin ngerjainnya.segera ya

f '(x) = 6x² - 2x + 5
maka
f(x) = ∫ 6x² - 2x + 5 dx
= 6x³/3 - 2x²/2 + 5x + c
= 3x³ - x² + 5x + C
diketahui f(0) = 5, maka
f(0) = 3(0)³ - (0)² + 5(0) + C = 5
⇔ C = 5
Jadi
f(x) = 3x³ - x² + 5x + 5


f '(x) = 1/2x²
maka
f(x) = ∫ 1/2x² dx
= ∫ 2x⁻² dx
= 2x⁻¹/-1 + C
= -2/x + C
diketahui f(3) = 7/6
f(3) = -2/(3) + C = 7/6
          C = 7/6 + 2/3
          C = 7/6 + 4/6
          C = 11/6
Jadi f(x) = -2/x + 11/6


f '(x) = 2(6x² - 1)
maka
f(x) = ∫ 2(6x² - 1) dx
= ∫ 12x² - 2 dx
= 12x³/3 - 2x + C
= 4x³ - 2x + C
diketahui f(0) = 5 dan f(1) = 2
f(0) = 4(0)³ - 2(0) + C = 5
⇔ C = 5
f(1) = 4(1)³ - 2(1) + C = 2
⇔ C = 0
Jadi
f(x) = 4x³ - 2x + 5 (jika c=5)

f(x) = 4x³ - 2x  (jika c = 0)