Problem 41 - Calculus Single Variable

Question

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{-5}^{2} \ln (x+5) d x\)

Step-by-Step Solution

Answer

The integral \(\int_{-5}^{2} \ln(x+5) \, dx\) is divergent.

Step 1: Recognize the Form of the Integral

The integral \( \int_{-5}^{2} \ln(x+5) \, dx \) is improper because the lower limit of integration, \( x = -5 \), causes the logarithmic function \( \ln(x+5) \) to become undefined due to the logarithm of zero.

Step 2: Establish the Improper Integral as a Limit

To handle the point where the integrand becomes undefined, rewrite the integral as a limit: \[\int_{-5}^{2} \ln(x+5) \, dx = \lim_{a \to -5^+} \int_{a}^{2} \ln(x+5) \, dx.\] This approach isolates the problematic behavior at \( x = -5 \).

Step 3: Compute the Antiderivative

The antiderivative of \( \ln(x+5) \) can be found using integration by parts. Choose \( u = \ln(x+5) \) and \( dv = dx \). Then \( du = \frac{1}{x+5} \ dx \) and \( v = x \). Thus:\[ \int \ln(x+5) \, dx = x \ln(x+5) - \int \frac{x}{x+5} \, dx. \]

Step 4: Simplify the Inner Integral

The integral \( \int \frac{x}{x+5} \, dx \) requires decomposition. Note that:\[ \frac{x}{x+5} = 1 - \frac{5}{x+5}. \]Thus, the integral becomes:\[ \int \frac{x}{x+5} \, dx = \int \left( 1 - \frac{5}{x+5} \right) dx = x - 5 \ln|x+5| + C. \]

Step 5: Calculate the Definite Integral

Substitute back the result into the original antiderivative, we have: \[ \int \ln(x+5) \, dx = x \ln(x+5) - x + 5 \ln|x+5| + C. \]Now evaluate the definite integral:\[ \lim_{a \to -5^+} \left[ x \ln(x+5) - x + 5 \ln|x+5| \right]_{a}^{2}. \]

Step 6: Evaluate at Upper Limit

Plug \(x = 2\) into the antiderivative:\[ 2 \ln(2+5) - 2 + 5 \ln|2+5| = 2 \ln 7 - 2 + 5 \ln 7. \] This simplifies to \(7 \ln 7 - 2\).

Step 7: Evaluate at Lower Limit

Evaluate the limit as \(a\) approaches \(-5^+\) from the positive side:\[ \lim_{a \to -5^+} \left[ a \ln(a+5) - a + 5 \ln|a+5| \right]. \]Since \( a \ln (a+5) \to 0 \) and \( -a \to 5 \), we focus on:\[ 0 + 5 \lim_{a \to -5^+} \ln(a+5). \] This goes to \(-\infty\) as \(a \to -5^+\).

Step 8: Conclusion

Since the lower limit evaluation results in \(-\infty\), this indicates the integral diverges at \(x = -5\). Thus, the original integral is divergent.

Key Concepts

Convergence and Divergence

Improper integrals are integrals that have infinite limits or discontinuities within the interval of integration. A key aspect of these integrals is determining whether they are convergent or divergent.

• Convergent Integral: This occurs when the limit of the integral exists and is finite as it approaches its bounds or points of discontinuity.
• Divergent Integral: This is when the limit does not exist or is infinite, meaning the area under the curve extends indefinitely.

For the improper integral \(\int_{-5}^{2} \ln(x+5) \, dx\), the function becomes undefined at \(x = -5\) due to the logarithm's behavior. When evaluating such integrals, it's crucial to analyze the behavior of the integrand at these problematic points by considering limits.In this exercise, the conclusion about divergence is based on the finding that the limit at \(-5^+\) reaches \(-\infty\). This indicates an infinite area under the curve from \(-5\) to 2, resulting in a divergent integral.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation, used to integrate products of functions. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]To apply this method, you must identify parts of the function to set as \(u\) and \(dv\), which then allows for easier integration. In this context:

• Choose: \(u = \ln(x+5)\), thus \(du = \frac{1}{x+5} \, dx\).
• Set: \(dv = dx\), leading to \(v = x\).

Applying integration by parts gives:\[\int \ln(x+5) \, dx = x \ln(x+5) - \int \frac{x}{x+5} \, dx\]The subsequent integral, \(\int \frac{x}{x+5} \, dx\), may require additional manipulation, such as partial fraction decomposition, to simplify and solve. The integration by parts technique here simplifies the original expression, making the integral more manageable.

Limits in Integrals

When dealing with improper integrals, particularly those with discontinuities or undefined points within the integration range, limits play a crucial role. In these cases, the integral is expressed in terms of a limit to handle the problematic point.To understand why limits are important, consider when the function approaches a point where it becomes undefined, like in the integral \(\int_{-5}^{2} \ln(x+5) \, dx\). The logarithmic function causes the integrand to become undefined at \(x = -5\), as \(\ln(0)\) is not defined.To address this:

• Rewrite: The integral as a limit: \[\lim_{a \to -5^+} \int_{a}^{2} \ln(x+5) \, dx\]
• Analyze: Evaluate the behavior of the integral as \(a\) approaches \(-5\) from the right, ensuring the function remains within the valid domain.

Often, such limits will reveal if the improper integral is convergent or divergent, as in this case where the resulting limit was \(-\infty\), indicating divergence. This technique helps circumvent issues at boundaries or points of discontinuity, ensuring a valid evaluation of the integral's behavior.