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.

R is symmetric because if (x,y) is in R, then (y,x) is also in R. It is not reflexive or transitive.

5 kilometers is equal to 5000 meters.

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.

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

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 coefficient of x^2 is C(4,2) * (3)^2 * (-4)^2 = 6 * 9 * 16 = 864.

The coefficient of x^2 is C(6,2) * (3)^2 * (-4)^4 = 15 * 9 * 256 = 34560.

The coefficient of x^2 is C(6,2)(-2)^4 = 15 * 16 = 240.

The lines are perpendicular if 2h = a + b, which leads to h^2 = -ab.

The lines are parallel if the discriminant of the quadratic equation is zero, which leads to the condition h^2 = ab.

For the lines to be coincident, the discriminant must be zero, i.e., b^2 - 4ac = 0.

The lines are perpendicular if the condition a + b = 0 holds true.

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.