Integral (5x + 1)(3x + 5) batas atas 2 batas bawah -1

Penyelesaian:

₋₁ഽ² (5x + 1)(3x + 5) dx
= ₋₁ഽ² (15x² + 28x + 5) dx
= [15 × ⅓x³ + 28 × ½x² + 5x]²
                                                 ⁻¹
= [5x³ + 14x² + 5x]²
                                ⁻¹
= (5 × 2³ + 14 × 2² + 5 × 2) – (5 × (-1)³ + 14 × (-1)² + 5 × (-1))
= (40 + 56 + 10) – (-5 + 14 – 5)
= 106 – 4
= 102
Kaprikornus, hasil ₋₁ഽ² (5x + 1)(3x + 5) dx = 102.
  Akar-akar persamaan kuadrat x² + (a – 1)x + 2 = 0 adalah α dan β.