∫(0 to 1) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 1 = (1/3 + 1) - (0) = 4/3.

∫(0 to 2) (x^2 + 1) dx = [x^3/3 + x] from 0 to 2 = (8/3 + 2) - (0) = 4.

∫(0 to π) sin(x) dx = [-cos(x)] from 0 to π = [1 - (-1)] = 2.

∫(1 to 3) (x^2 - 2x + 1) dx = [(x^3/3 - x^2 + x)] from 1 to 3 = (9/3 - 9 + 3) - (1/3 - 1 + 1) = 0.

∫(1 to 4) (x^3) dx = [x^4/4] from 1 to 4 = (256/4 - 1/4) = 63.

The integral ∫(2x + 1)dx = x^2 + x. Evaluating from 0 to 2 gives (2^2 + 2) - (0 + 0) = 4.

To find the minimum, set f'(x) = e^x + 2x = 0. The minimum occurs at x = 0.

Set f'(x) = e^x - 2x = 0. The solution is approximately x = 1.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting 6x - 12 = 0 gives x = 2.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting 6x - 12 = 0 gives x = 2.

Setting f'(x) = 3x^2 - 12x + 9 = 0 gives x = 1, 3. Testing shows x = 2 is a local minimum.

To find inflection points, set f''(x) = 0. f''(x) = 6x - 12. Setting it to zero gives x = 2.

By substituting x = 1 into the equation, we find that it satisfies the equation, hence 1 is a root.

The roots can be found using the Rational Root Theorem and synthetic division, confirming that all roots are real.

The mode is 4, as it appears 3 times, more than any other number.

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3. Variance = [(1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)²] / 5 = 2.

Mean = (10 + 12 + 14 + 16 + 18) / 5 = 14. Variance = [(10-14)² + (12-14)² + (14-14)² + (16-14)² + (18-14)²] / 5 = 8.

Mean = (10 + 12 + 14 + 16) / 4 = 13. Variance = [(10-13)² + (12-13)² + (14-13)² + (16-13)²] / 4 = 2.

Mean = (2 + 4 + 6 + 8 + 10) / 5 = 6. Variance = [(2-6)² + (4-6)² + (6-6)² + (8-6)² + (10-6)²] / 5 = 8.

All values are the same, so variance = 0.

Mean = 9. Variance = [(5-9)² + (7-9)² + (9-9)² + (11-9)² + (13-9)²] / 5 = 8.

Arranging the numbers: 5, 8, 12, 15, 20. The median is the middle number, which is 12.

The mode is 9, as it appears most frequently (three times) in the data set.

The discriminant is 2^2 - 4(1)(1) = 0, indicating that the roots are real and equal.

The discriminant must be non-negative: 4^2 - 4*1*k >= 0 leads to 16 - 4k >= 0, thus k <= 4.

The condition for no real roots is that the discriminant must be less than zero: 6^2 - 4*1*k < 0 => 36 < 4k => k > 9.

The equation can be factored as (x-1)^3 = 0, which has one real root (x = 1) with multiplicity 3.

By substituting x = 2 into the equation, we find that it satisfies the equation, thus x = 2 is a root.

The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.