Решите проверочную по алгебре пожалуйста(желательно на листочке)

1)
∫(-2, 3)(4x³ - 3x² + 2x + 1)dx = x⁴ - x³ + x² + x | (-2, 3) = (3⁴ - 3³ + 3² + 3) - ((-2)⁴ - (-2)³ + (-2)² + (-2)) = (81 - 27 + 9 + 3) - (16 + 8 + 4 - 2) = 66 - 26 = 40

2)
∫(-1, 2)(4/3x³ - 3/4x² + 5)dx = 1/3x⁴ - 1/4x³ + 5x | (-1, 2) = (1/3 * 2⁴ - 1/4 * 2³ + 5 * 2) - (1/3 * (-1)⁴ - 1/4 * (-1)³ + 5 * (-1)) = (16/3 - 8/4 + 10) - (1/3 + 1/4 - 5) = 16/3 - 2 + 10 - 1/3 - 1/4 + 5 = 5 + 15/3 - 1/4 = 5 + 5 - 1/4 = 10 - 1/4 = 39/4 = 9,75

3)
y = x² - 4 и y = 0
x² - 4 = 0
x² = 4
x1 = 2; x2 = -2
∫(-2, 2)(0 - (x² - 4))dx = ∫(-2, 2)(-x² + 4)dx = -1/3x³ + 4x | (-2, 2) = (-1/3 * 2³ + 4 * 2) - (-1/3 * (-2)³ + 4 * (-2)) = (-8/3 + 8) - (8/3 - 8) = -8/3 + 8 - 8/3 + 8 = 16 - 16/3 = 32/3 ≈ 10,67