(b) ₁₂ (x − 1)(x + 2) de L dx - -2

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Chapter1: Functions And Models
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Write the improper integral attached as a limit or sum of limits. Do not evaluate the integrals or limits.

### Integral Evaluation Example

**Problem (b):**

Evaluate the integral:

\[ \int_{-2}^{0} \frac{1}{(x-1)(x+2)} \, dx \]

**Explanation:**

This problem requires us to evaluate the definite integral of the given function from \( x = -2 \) to \( x = 0 \).

The integrand is a rational function given by:

\[ \frac{1}{(x-1)(x+2)} \]

To solve this integral, we typically use partial fraction decomposition to simplify the integrand into a sum of simpler fractions, which can then be integrated separately.

**Steps:**

1. **Partial Fraction Decomposition**:
   
   Express \( \frac{1}{(x-1)(x+2)} \) as:

   \[ \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \]

   To find A and B, solve for the constants by clearing the denominators:

   \[ 1 = A(x+2) + B(x-1) \]

   Let \( x = 1 \):

   \[ 1 = A(1+2) + B(1-1) \]
   \[ 1 = 3A \]
   \[ A = \frac{1}{3} \]

   Let \( x = -2 \):

   \[ 1 = A(-2+2) + B(-2-1) \]
   \[ 1 = -3B \]
   \[ B = -\frac{1}{3} \]

   Thus, we have:

   \[ \frac{1}{(x-1)(x+2)} = \frac{1/3}{x-1} - \frac{1/3}{x+2} \]

2. **Rewrite the Integral**:

   \[ \int_{-2}^{0} \left( \frac{1/3}{x-1} - \frac{1/3}{x+2} \right) \, dx \]

   Factor out the constant \( \frac{1}{3} \):

   \[ \frac{1}{3} \int_{-2}^{0} \left(\frac{1}{x-1}
Transcribed Image Text:### Integral Evaluation Example **Problem (b):** Evaluate the integral: \[ \int_{-2}^{0} \frac{1}{(x-1)(x+2)} \, dx \] **Explanation:** This problem requires us to evaluate the definite integral of the given function from \( x = -2 \) to \( x = 0 \). The integrand is a rational function given by: \[ \frac{1}{(x-1)(x+2)} \] To solve this integral, we typically use partial fraction decomposition to simplify the integrand into a sum of simpler fractions, which can then be integrated separately. **Steps:** 1. **Partial Fraction Decomposition**: Express \( \frac{1}{(x-1)(x+2)} \) as: \[ \frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \] To find A and B, solve for the constants by clearing the denominators: \[ 1 = A(x+2) + B(x-1) \] Let \( x = 1 \): \[ 1 = A(1+2) + B(1-1) \] \[ 1 = 3A \] \[ A = \frac{1}{3} \] Let \( x = -2 \): \[ 1 = A(-2+2) + B(-2-1) \] \[ 1 = -3B \] \[ B = -\frac{1}{3} \] Thus, we have: \[ \frac{1}{(x-1)(x+2)} = \frac{1/3}{x-1} - \frac{1/3}{x+2} \] 2. **Rewrite the Integral**: \[ \int_{-2}^{0} \left( \frac{1/3}{x-1} - \frac{1/3}{x+2} \right) \, dx \] Factor out the constant \( \frac{1}{3} \): \[ \frac{1}{3} \int_{-2}^{0} \left(\frac{1}{x-1}
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