The area is given by the integral from 0 to 1 of (x - x^2) dx. This evaluates to [x^2/2 - x^3/3] from 0 to 1 = (1/2 - 1/3) = 1/6 = 0.5.

The area is given by the integral from 0 to 2 of (2x - x^2) dx. This evaluates to [x^2 - x^3/3] from 0 to 2 = (4 - 8/3) = 4/3.

The area is given by the integral from 0 to 2 of (4 - x^2) dx. This evaluates to [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 16/3.

The area under the curve y = cos(x) from x = 0 to x = π/2 is given by ∫(from 0 to π/2) cos(x) dx = [sin(x)] from 0 to π/2 = 1 - 0 = 1.

The area under the curve is given by ∫(from 0 to 2) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 2 = (8/3 + 4) = 20/3.

The area under the curve y = x^4 from x = 0 to x = 2 is given by ∫(from 0 to 2) x^4 dx = [x^5/5] from 0 to 2 = (32/5) - 0 = 32/5.

Using the chain rule, f'(x) = d/dx(e^(2x)) = 2e^(2x).

Using the product rule, f'(x) = d/dx(x^2 * e^x) = (x^2 + 2x)e^x.

The integral of x^2 is (1/3)x^3, the integral of 2x is x^2, and the integral of 1 is x. Thus, ∫ (x^2 + 2x + 1) dx = (1/3)x^3 + x^2 + x + C.

The integrand simplifies to x + 1. Therefore, ∫ (x + 1) dx = (1/2)x^2 + x + C.

The integral of x^3 is (1/4)x^4 and the integral of -4x is -2x^2. Therefore, ∫ (x^3 - 4x) dx = (1/4)x^4 - 2x^2 + C.

Using the identity cos^2(x) = (1 + cos(2x))/2, we find that ∫ cos^2(x) dx = (1/2)x + (1/4)sin(2x) + C.

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

Using the identity 1 - cos(x) = 2sin^2(x/2), we have lim (x -> 0) (2sin^2(x/2))/(x^2) = 1.

Using the definition of the derivative of e^x at x = 0, we find that lim (x -> 0) (e^x - 1)/x = e^0 = 1.

Using the standard limit lim (x -> 0) (tan(kx)/x) = k, we have k = 3, so the limit is 3.

This is an indeterminate form (0/0). Factor the numerator: (x-1)(x+1)/(x-1) = x + 1. Thus, lim (x -> 1) (x + 1) = 2.

Factoring gives (x - 1)(x + 1)/(x - 1)^2 = (x + 1)/(x - 1). Thus, lim (x -> 1) (x + 1)/(x - 1) = 2.

Factoring gives (x - 1)(x^2 + x + 1)/(x - 1). Canceling (x - 1) gives lim (x -> 1) (x^2 + x + 1) = 3.

Factoring gives (x(x - 2))/(x - 2), canceling gives lim (x -> 2) x = 2.

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

The integral evaluates to [ln(x)] from 0 to 1 = ln(1) - ln(0) which diverges.

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

The integral evaluates to [x^4 - (2/3)x^3 + (1/2)x^2] from 0 to 1 = (1 - 2/3 + 1/2) = 1/6.

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

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

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

Using integration by parts, the integral evaluates to (2/e - 1/e) = 1/e.

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

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