The integral evaluates to [x^4/4 - 2x^2 + 4x] from 0 to 1 = (1/4 - 2 + 4) = (1/4 + 2) = 9/4.

The integral evaluates to [x^5/5 + (2/3)x^3] from 0 to 1 = (1/5 + 2/3) = 11/15.

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

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

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

The integral evaluates to [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 4 + 1 = 5.

The integral evaluates to [x^3 - x^2 + x] from 1 to 2 = (8 - 4 + 2) - (1 - 1 + 1) = 5.

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

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

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

The integral evaluates to [x^5/5 + (1/2)x^4 + (1/3)x^3] from 0 to 1 = 1/5 + 1/2 + 1/3 = 31/30.

∫_0^1 (x^4 + 2x^3) dx = [x^5/5 + (1/2)x^4] from 0 to 1 = (1/5 + 1/2) = 7/10.

∫_0^1 x^4 dx = [x^5/5] from 0 to 1 = 1/5.

Using the identity sin(2x) = 2sin(x)cos(x), the integral becomes (1/2)∫_0^π sin(2x) dx = 0.

∫_0^π sin(x) dx = [-cos(x)] from 0 to π = -(-1 - 1) = 2.

The integral evaluates to [x/2 + sin(2x)/4] from 0 to π/2 = π/4.

Solving gives cos^2(x) = 1/3, so x = π/3, 2π/3, and their equivalents.

The solution is x = π/6 + 2nπ.

Factoring gives (sin(x) - 2)(sin(x) + 1) = 0, so sin(x) = 2 (not possible) or sin(x) = -1, giving x = 3π/2.

Factoring gives sin(x)(sin(x) - 1) = 0, so x = 0 and x = π.

Rearranging gives sin(x) = 1/3, which has solutions in the specified interval.

The solution is x = π/2.

Factoring gives sin(x)(sin(x) - 1) = 0, so x = 0 and x = π.

Weighted mean = (10*1 + 20*2 + 30*3) / (1 + 2 + 3) = (10 + 40 + 90) / 6 = 140 / 6 = 23.33.

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

The vertex occurs at x = -b/(2a) = 4/2 = 2. This is where the local minimum occurs.

The vertex occurs at x = -b/(2a) = 4/2 = 2, which is the x-coordinate of the minimum point.

Set x = 0: -2y = 10 => y = -5. The y-intercept is (0, -5).

The electric field due to a charged plane sheet is directly proportional to the surface charge density. Therefore, if σ is doubled, E also doubles.

The electric field inside a uniformly charged sphere is zero.