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

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

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

The integral evaluates to [x^3/3 - 4x] from 1 to 3 = (27/3 - 12) - (1/3 - 4) = 2.

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

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

∫_0^1 (x^3 + 2x^2) dx = [x^4/4 + 2x^3/3] from 0 to 1 = (1/4 + 2/3) = 11/12.

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

∫_0^1 (x^4 - 2x^2 + 1) dx = [x^5/5 - (2/3)x^3 + x] from 0 to 1 = (1/5 - 2/3 + 1) = 1/15.

The integral evaluates to [x^5/5] from 0 to 1 = 1/5.

∫_0^π/2 cos^2(x) dx = π/4.

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

The integral evaluates to [x^3 - 4x] from 1 to 2 = (8 - 8) - (1 - 4) = 3.

∫_1^2 (3x^2 - 4x + 1) dx = [x^3 - 2x^2 + x] from 1 to 2 = (8 - 8 + 2) - (1 - 2 + 1) = 1.

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

The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = 16/3.

The area between the curves y = x^2 and y = 4 is given by ∫(from 0 to 2) (4 - x^2) dx = [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 4/3.

The area between the curves is given by ∫(from 0 to 1) (x - x^3) dx = [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.

The area is given by the integral from 0 to 1 of e^x dx. This evaluates to [e^x] from 0 to 1 = e - 1.

The area under the curve is given by ∫(from 0 to 3) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 3 = (27/3 + 9) = 18.

The area under the curve y = x^2 from 0 to 2 is given by the integral ∫(from 0 to 2) x^2 dx = [x^3/3] from 0 to 2 = (2^3/3) - (0^3/3) = 8/3.

Area = ∫ from 0 to 3 of x^2 dx = [1/3 * x^3] from 0 to 3 = 9.

The area is given by the integral ∫ (x^2) dx from 1 to 3. This evaluates to [x^3/3] from 1 to 3 = (27/3 - 1/3) = 26/3.

The area under the curve y = x^4 from 0 to 1 is given by ∫(from 0 to 1) x^4 dx = [x^5/5] from 0 to 1 = 1/5.

The area is given by the integral from 0 to 2 of x^4 dx. This evaluates to [x^5/5] from 0 to 2 = (32/5) = 16.

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x^2 = 1, so x = 1 or x = -1. f(1) = 0, f(-1) = 4. The point is (1, 0).

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. The point is (1, 0).

f'(x) = 6x - 12. Setting f'(x) = 0 gives x = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

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

f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x^2(12x - 24) = 0, so x = 0 or x = 2. f(2) = 3(2^4) - 8(2^3) + 6 = -2.