Question

Give some examples of analytical methods for evaluating integrals.

Short answer

Question: Give examples of four different analytical methods for evaluating integrals and briefly describe each method. Answer: The four analytical methods for evaluating integrals include: 1. Direct Integration: This method involves finding the exact value of an integral using anti-derivatives or by reversing the differentiation process. It is used when the function is simple and can be directly integrated. 2. Substitution: This technique is used when the integral can be transformed into a simpler one by substituting a new variable for a part of the integrand. The chain rule is then used to rewrite the integral in terms of the new variable. 3. Integration by Parts: This technique is used for integrating products of functions and is based on the product rule for differentiation. It involves choosing appropriate functions u and v from the given integral and applying the integration by parts formula. 4. Partial Fractions: This technique is used to break up fractions of polynomials into simpler fractions to make them easier to integrate. The main idea is to express a given fraction as the sum of simpler fractions with unknown coefficients, which can then be integrated separately.

Step-by-step solution

1. Direct Integration

Direct integration involves finding the exact value of an integral using anti-derivatives or by reversing the differentiation process. If the function is simple and can be directly integrated, this method is the easiest and most straightforward way of solving an integral. For example, consider the integral: \[ \int x^2 dx \] The anti-derivative of \(x^2\) is \(\frac{1}{3}x^3\), so the result of the direct integration is: \[ \int x^2 dx = \frac{1}{3}x^3 + C \] where C is the constant of integration.

2. Substitution

Substitution is a technique used when the integral can be transformed into a simpler one by substituting a new variable for a part of the integrand. The chain rule is then used to rewrite the integral in terms of the new variable. For example, consider the integral: \[ \int x e^{x^2} dx \] With the substitution \(u = x^2\), we have \(du = 2x dx\). The integral can be rewritten as: \[ \int \frac{1}{2} e^u du \] Direct integration yields: \[ \int \frac{1}{2} e^u du = \frac{1}{2} e^u + C \] Now, substitute back \(u = x^2\): \[ \frac{1}{2} e^{x^2} + C \]

3. Integration by Parts

Integration by parts is a technique used for integrating products of functions. It is based on the product rule for differentiation. Given the integral \(\int u dv\), integration by parts states that: \[ \int u dv = uv - \int v du \] For example, consider the integral: \[ \int x e^x dx \] Choose \(u = x\) and \(dv = e^x dx\), then \(du = dx\) and \(v = e^x\). Applying integration by parts: \[ \int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C \]

4. Partial Fractions

Partial fractions are used to break up fractions of polynomials into simpler fractions to make them easier to integrate. The main idea is to express a given fraction as the sum of simpler fractions with unknown coefficients. For example, consider the integral: \[ \int \frac{1}{x^2 - 1} dx \] First, express the fraction as partial fractions: \[ \frac{1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1} \] We need to find A and B. After solving for A and B, we get \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\), so the integral becomes: \[ \int \frac{1}{x^2 - 1} dx = \int \frac{1}{2} \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) dx \] Now, direct integration yields: \[ \frac{1}{2} \int \frac{1}{x - 1} dx - \frac{1}{2} \int \frac{1}{x + 1} dx = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x + 1| + C \]