Analysis and Calculus

We can use L'Hopital's rule to evaluate the limit. Taking the derivative of the numerator and denominator, we get (f'(x) = \frac{2x}{1} = 2x) and (g'(x) = \frac{1}{1} = 1). Substituting (x = 2), we get (f'(2) = 4) and (g'(2) = 1). Therefore, the limit of (f(x)) as (x) approaches (2) is (\frac{4}{1} = 4).

Using the power rule of differentiation, we get (f'(x) = \frac{d}{dx}(x^3 - 2x^2 + 3x - 4) = 3x^2 - 2(2x) + 3(1) - 0 = 3x^2 - 4x + 3).

Using the power rule of integration, we get (\int_0^2 x^2 dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}).

The area under the curve can be found by evaluating the integral (\int_0^2 x^2 dx). Using the power rule of integration, we get (\int_0^2 x^2 dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}).

To find the equation of the tangent line, we need to find the slope of the curve at the point ((1, 0)). The slope is given by the derivative of the function, which is (f'(x) = 3x^2 - 4x + 3). Substituting (x = 1), we get (f'(1) = 3(1)^2 - 4(1) + 3 = 2). Therefore, the equation of the tangent line is (y - 0 = 2(x - 1)), which simplifies to (y = x - 1).

To find the volume of the solid, we need to use the formula (V = \pi\int_a^b [f(x)]^2 dx), where (f(x)) is the radius of the cross-section and ([a, b]) is the interval of integration. In this case, (f(x) = 4 - x^2) and ([a, b] = [0, 2]). Substituting these values into the formula, we get (V = \pi\int_0^2 (4 - x^2)^2 dx). Evaluating this integral, we get (V = \frac{64\pi}{3}).

To find the general solution, we need to integrate both sides of the differential equation with respect to (x). Integrating (\frac{dy}{dx} = 2x + 1) gives (y = \int (2x + 1) dx = \int 2x dx + \int 1 dx = x^2 + x + C), where (C) is the constant of integration.

The improper integral (\int_0^\infty \frac{1}{x} dx) diverges to (\infty). This is because the function (f(x) = \frac{1}{x}) has an infinite discontinuity at (x = 0), and the integral (\int_0^a \frac{1}{x} dx) diverges to (\infty) as (a) approaches (0).

To find the area of the region, we need to find the points of intersection of the two curves. Setting (y = x^2) and (y = 2x + 1) equal to each other, we get (x^2 = 2x + 1). Solving for (x), we get (x = -1) and (x = 1). Therefore, the area of the region is (\int_{-1}^1 (2x + 1 - x^2) dx = \left[x^2 + x - \frac{x^3}{3}\right]_{-1}^1 = \frac{3}{2}).

Using the chain rule, we get (f'(x) = \frac{d}{dx}[\sin(x^2 + 1)]\frac{d}{dx}[x^2 + 1] = \cos(x^2 + 1)(2x) = 2x\cos(x^2 + 1)).

Using the partial fraction decomposition, we can write (\frac{1}{x^2 - 4} = \frac{1}{2(x - 2)} + \frac{1}{2(x + 2)}). Integrating each term, we get (\int \frac{1}{x^2 - 4} dx = \frac{1}{2}\ln|x - 2| + \frac{1}{2}\ln|x + 2| + C), where (C) is the constant of integration.

To find the equation of the tangent line, we need to find the slope of the curve at the point ((2, 1)). The slope is given by the derivative of the function, which is (f'(x) = x^2 - 4x + 4). Substituting (x = 2), we get (f'(2) = 2^2 - 4(2) + 4 = 0). Therefore, the equation of the tangent line is (y - 1 = 0(x - 2)), which simplifies to (y = 3x - 1).

To find the volume of the solid, we need to use the formula (V = \pi\int_a^b [f(x)]^2 dx), where (f(x)) is the radius of the cross-section and ([a, b]) is the interval of integration. In this case, (f(x) = 4 - x^2 - x^2 = 4 - 2x^2) and ([a, b] = [-2, 2]). Substituting these values into the formula, we get (V = \pi\int_{-2}^2 (4 - 2x^2)^2 dx). Evaluating this integral, we get (V = \frac{64\pi}{3}).

The characteristic equation of the differential equation is (r^2 + 4 = 0), which has roots (r = \pm 2i). Therefore, the general solution is (y = A\cos(2x) + B\sin(2x)), where (A) and (B) are constants.

To find the area of the surface, we need to use the formula (S = 2\pi\int_a^b f(x)\sqrt{1 + [f'(x)]^2} dx), where (f(x)) is the function and ([a, b]) is the interval of integration. In this case, (f(x) = x^2), (f'(x) = 2x), and ([a, b] = [0, 2]). Substituting these values into the formula, we get (S = 2\pi\int_0^2 x^2\sqrt{1 + (2x)^2} dx). Evaluating this integral, we get (S = \frac{64\pi}{3}).