Integration and Definite Integrals

To evaluate the integral, we can use the power rule of integration: (\int_{0}^{1} x^2 dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}).

To find the indefinite integral, we can use partial fraction decomposition to rewrite the integrand as (\frac{x^2 + 2x - 3}{x - 1} = x + 3 + \frac{1}{x - 1}). Then, we can integrate each term separately: (\int (x + 3 + \frac{1}{x - 1}) dx = \frac{x^2}{2} + 3x + \ln|x - 1| + C), where (C) is the constant of integration.

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 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).

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.

The symbol ∫ is used to represent the integral of a function.

The integral of a function f(x) is given by the formula F(x) = ∫f(x) dx. Substituting the value of f(x), we get F(x) = ∫(2x^3 - 3x^2 + 4x - 5) dx. Integrating each term separately, we get F(x) = (2/4)x^4 - (3/3)x^3 + (4/2)x^2 - 5x + C. Simplifying this equation, we get F(x) = x^4 - x^3 + 2x^2 - 5x + C, where C is the constant of integration.

To find the indefinite integral of (f(x)), we apply the power rule of integration. The indefinite integral of (x^n) is (\frac{x^{n+1}}{n+1} + C), where (C) is the constant of integration.

To evaluate the definite integral, we first find the indefinite integral of the integrand (3x^2 - 2x + 1) using the power rule and the constant rule of integration. Then, we evaluate the indefinite integral at the upper and lower limits of integration and subtract the result at the lower limit from the result at the upper limit.

To solve this integral using integration by substitution, let (u = 3x). Then, (du = 3 dx). Substituting (u) and (du) into the integral, we get (\int \sin(3x) dx = \int \sin(u) \frac{1}{3} du). Now, we can integrate (\sin(u)) using the power rule of integration.

To solve this integral using integration by parts, let (u = e^{2x}) and (dv = dx). Then, (du = 2e^{2x} dx) and (v = x). Substituting (u), (du), (v), and (dv) into the integration by parts formula, we get (\int e^{2x} dx = xe^{2x} - \int 2xe^{2x} dx). Now, we can solve the remaining integral using the power rule of integration.

An improper integral is an integral where the interval of integration is infinite or contains a point where the integrand is undefined. To evaluate an improper integral, we need to take the limit of the definite integral as the upper or lower limit of integration approaches infinity or negative infinity, or as the integrand approaches infinity or negative infinity at a point in the interval. In this case, the improper integral (\int_1^\infty \frac{1}{x^2} dx) can be evaluated as (\lim_{x \to \infty} \int_1^x \frac{1}{x^2} dx).