The integral evaluates to [x^3/3 + 2x^2 + 4x] from 0 to 1 = (1/3 + 2 + 4) = 25/3.

The integral evaluates to [x^5/5 - 2x^3/3 + x] from 0 to 1 = (1/5 - 2/3 + 1) = (15/15 - 10/15 + 3/15) = 8/15.

The integral evaluates to [x^5/5 - (2/4)x^4 + (1/3)x^3] from 0 to 1 = (1/5 - 1/2 + 1/3) = 1/30.

The integral evaluates to [x^4/4 - x^3 + 4x] from 0 to 2 = (4 - 8 + 8) - 0 = 4.

The integral evaluates to [-cos(x)] from 0 to π = [1 - (-1)] = 2.

Using the identity sin(2x) = 2sin(x)cos(x), the integral becomes 1/2 ∫ from 0 to π/2 of sin(2x) dx = 1/2 [-1/2 cos(2x)] from 0 to π/2 = 1/2 [0 - (-1/2)] = 1/4.

Using the identity sin^2(x) = (1 - cos(2x))/2, the integral evaluates to π/4.

The integral evaluates to [x^2 + x] from 1 to 3 = (9 + 3) - (1 + 1) = 10.

∫_0^1 (4x^3 - 3x^2 + 2) dx = [x^4 - x^3 + 2x] from 0 to 1 = (1 - 1 + 2) - (0) = 2.

∫_0^1 e^x dx = [e^x] from 0 to 1 = e - 1.

The integral evaluates to [x^4/4 - 2x^3/3 + x^2/2] from 0 to 1 = (1/4 - 2/3 + 1/2) = 1/12.

∫_0^π/2 cos^2(x) dx = π/4.

∫_1^e ln(x) dx = [x ln(x) - x] from 1 to e = (e - e) - (1 - 1) = 1.

Using integration by parts, the integral evaluates to 1.

f(x) is a polynomial function, which is differentiable everywhere, including at x = 1.

The left limit is 0 and the right limit is undefined. Thus, f(x) is not continuous at x = 0.

At x = 1, f(1) = 3, but lim x->1- f(x) = 1 and lim x->1+ f(x) = 2. Thus, it is not continuous.

Both sides equal 2 at x = 1, hence it is continuous.

The left-hand derivative is -1 and the right-hand derivative is 1. Since they are not equal, f(x) is not differentiable at x = 1.

The area is given by the integral from 0 to 1 of (x - x^3) dx. This evaluates to [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.

The area enclosed is found by integrating from -2 to 2: ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 16/3.

The area under the curve y = 1/x from x = 1 to x = 2 is given by ∫(from 1 to 2) (1/x) dx = [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).

The area under the curve y = e^x from 0 to 1 is given by ∫(from 0 to 1) e^x dx = [e^x] from 0 to 1 = e - 1.

The limit as x approaches 0 does not exist, hence f(x) is not continuous at x = 0.

The left limit is 0, the right limit is 2, and f(1) = 3. Thus, it is discontinuous.

The left limit as x approaches 1 is 1, and the right limit is also 1. Thus, f(1) = 1, making it continuous.

Setting f'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so critical points are x = -1 and x = 1.

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x = 0 and x = 2. Critical points are (0, 4) and (2, 1).

Setting f'(x) = 0 gives critical points at x = 0 and x = 3.

Setting f'(x) = 0 gives critical points at x = 1 and x = 2.