/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Converge. Evaluate the integrals... [FREE SOLUTION] | 91Ó°ÊÓ

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Converge. Evaluate the integrals without using tables. $$\int_{-1}^{\infty} \frac{d \theta}{\theta^{2}+5 \theta+6}$$

Short Answer

Expert verified
The integral converges to \(-\ln(2)\).

Step by step solution

01

Identify the integrand

The integrand is \( \frac{1}{\theta^2+5\theta+6} \). This is a rational function with a quadratic expression in the denominator.
02

Factor the Denominator

To simplify the integrand, factor the quadratic expression in the denominator. \( \theta^2 + 5\theta + 6 = (\theta + 2)(\theta + 3) \). Thus, \( \frac{1}{\theta^2+5\theta+6} = \frac{1}{(\theta+2)(\theta+3)} \).
03

Partial Fraction Decomposition

Express the integrand as a sum of partial fractions: \( \frac{1}{(\theta+2)(\theta+3)} = \frac{A}{\theta+2} + \frac{B}{\theta+3} \). Solve for \( A \) and \( B \) by multiplying both sides by \((\theta+2)(\theta+3)\) and equating coefficients.
04

Solve for the Coefficients

Set up the equation \((A(\theta+3) + B(\theta+2) = 1)\) and solve for \( A \) and \( B \). This gives \( A = 1 \) and \( B = -1 \), resulting in \( \frac{1}{(\theta+2)(\theta+3)} = \frac{1}{\theta+2} - \frac{1}{\theta+3} \).
05

Solve the Integrals

Integrate each term separately: \[ \int \frac{1}{\theta+2} \, d\theta - \int \frac{1}{\theta+3} \, d\theta = \ln|\theta+2| - \ln|\theta+3| + C \].
06

Apply Limits of Integration

Evaluate the improper integral \( \int_{-1}^{\infty} \frac{d\theta}{\theta^2 + 5\theta + 6} \). First, consider the limit as \( \theta \to \infty \), which gives zero for both terms as the logarithmic terms cancel out. For \( \theta = -1 \), \( \ln|-1+2| - \ln|-1+3| = \ln(1) - \ln(2) = -\ln(2) \).
07

Conclusion on Convergence

Since the limits at both bounds are finite, the integral converges. The value of the integral is \(-\ln(2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
When dealing with rational functions, particularly those with polynomial denominators like in our integral, partial fraction decomposition can be a super handy technique. It allows us to express a complex rational function as a sum of simpler fractions. This is especially useful for integration because integrating simple fractions is usually easier.
To apply partial fraction decomposition, start by factoring the denominator of the rational function. In this case, we have \(\frac{1}{\theta^2+5\theta+6}\) with a denominator that factors as \((\theta+2)(\theta+3)\).
Next, you express the function as a sum of fractions: \(\frac{A}{\theta+2} + \frac{B}{\theta+3}\). Find the coefficients \(A\) and \(B\) by clearing the denominators and solving the resulting equation to match the original numerator. Here, you get \(A = 1\) and \(B = -1\), simplifying our problem significantly.
By decomposing the fraction, you're just breaking down the problem into manageable parts.
Convergence
Evaluating an improper integral requires checking if the integral converges or diverges. Convergence implies that as the variable approaches a boundary or infinity, the integral still results in a finite value.
In the formula \(\int_{-1}^{\infty} \frac{d\theta}{\theta^2 + 5\theta + 6}\), we deal with both an infinite upper limit and a finite approach at \(\theta = -1\). For convergence, examine each limit separately.
  • At \(\theta = \infty\), both parts of the integral, \(\ln|\theta+2|\) and \(\ln|\theta+3|\), tend to zero as their differences vanish.
  • Approaching \(\theta = -1\), evaluate using the substitution resulting from the partial fraction decomposition. The integral \(-\ln(2)\) is finite, ensuring that this part also doesn’t blow up to infinity.
Thus, having finite limits at all boundaries finds convergence of the integral.
Rational Function
A rational function is one big fraction with polynomials in both the numerator and the denominator. In this case, the function is \(\frac{1}{\theta^2+5\theta+6}\), which exhibits classic rational function behavior due to the polynomials involved.
This function is stable except at points that make the denominator zero, known as poles or zeros. Here, those points are \(\theta = -2\) and \(\theta = -3\), where the function can potentially become undefined. It's crucial to consider these when evaluating integrals as they can affect convergence.
Rational functions can often be quite complex, but through techniques like factoring and partial fraction decomposition, they can become more tractable for analytical work such as integration.
Integration Techniques
Understanding various integration techniques helps solve different types of integrals that arise in calculus. For rational functions, break them down using methods like partial fraction decomposition before integrating.
Once decomposed, integrate each fraction separately. For instance, \(\int \frac{1}{\theta+2} \, d\theta\) becomes \(\ln|\theta+2|\), a standard result from integration rules involving natural logarithms.
For each fraction formed, apply basic integration rules, such as the natural log rule. Combine results from each simple integral to form the final answer. For an improper integral, ensure to evaluate it properly by considering its limits correctly.
  • Mastery of substitution techniques can also aid in tackling re-shaped integrals after decomposition.
  • Knowing when solutions might approach \(\pm \infty\) refines convergence conclusions further.
Altogether, combining these techniques allows for thorough and successful integration approaches.

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