3.6 Q-3

Question Statement

Evaluate the integral:
$\int_{-2}^{0} \frac{1}{(2x - 1)^2} , dx$

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Background and Explanation

This problem involves integration of a rational function. To solve such integrals, we often need to apply substitution or recognize standard integrals. In this case, we can use the standard formula for integrating a function of the form $\frac{1}{(ax + b)^n}$.

The steps include:

1. Simplifying the integral using substitution if needed.
2. Applying the appropriate power rule for integration.
3. Evaluating the resulting expression at the given limits.

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Solution

1. Rewrite the Integral:

   The integral is: $\int_{-2}^{0} \frac{1}{(2x - 1)^2} , dx$

   First, notice that the denominator is of the form $(2x - 1)^{-2}$, so we can use a substitution to simplify this.

2. Apply Substitution:

   Let: $u = 2x - 1$
   Then: $du = 2 dx \quad \text{or} \quad \frac{du}{2} = dx$

   Now substitute into the integral. The limits change as well:

   • When $x = -2$, $u = 2(-2) - 1 = -5$.
   • When $x = 0$, $u = 2(0) - 1 = -1$.

   Substituting into the integral: $\int_{-2}^{0} \frac{1}{(2x - 1)^2} , dx = \frac{1}{2} \int_{-5}^{-1} u^{-2} , du$

3. Integrate:

   Now, apply the power rule for integration: $\int u^{-2} , du = \frac{u^{-1}}{-1} = -u^{-1}$

   Therefore: $\frac{1}{2} \int_{-5}^{-1} u^{-2} , du = \frac{1}{2} \left[-\frac{1}{u}\right]_{-5}^{-1}$

4. Evaluate the Integral:

   Now evaluate at the limits $u = -1$ and $u = -5$: $\frac{1}{2} \left[-\frac{1}{-1} + \frac{1}{-5}\right] = \frac{1}{2} \left[1 - \left(-\frac{1}{5}\right)\right]$

   Simplify: $= \frac{1}{2} \left[1 + \frac{1}{5}\right] = \frac{1}{2} \times \frac{6}{5} = \frac{3}{5}$

Thus, the value of the integral is: $\boxed{\frac{3}{5}}$

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Key Formulas or Methods Used

• Substitution method for integrals:
  $u = 2x - 1 \quad \text{and} \quad du = 2dx$

• Power rule for integration:
  $\int u^n , du = \frac{u^{n+1}}{n+1}$

• Evaluating definite integrals at the upper and lower limits.

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Summary of Steps

1. Set up the integral with substitution:
   $u = 2x - 1 \quad \text{and} \quad du = 2dx$

2. Change the limits of integration according to $u$ and simplify the integral:
   $\frac{1}{2} \int_{-5}^{-1} u^{-2} , du$

3. Apply the power rule for integration:
   $\frac{1}{2} \left[-\frac{1}{u}\right]_{-5}^{-1}$

4. Evaluate the integral at the limits:
   $\frac{1}{2} \left[1 + \frac{1}{5}\right] = \frac{3}{5}$

Thus, the final result is $\frac{3}{5}$.