Problem 28 - University Calculus: Early Transcendentals

Question

Integrate \(f\) over the given region. Curved region \(f(s, t)=e^{s}\) Int over the region in the first quadrant of the st-plane that lies above the curve \(s=\ln t\) from \(t=1\) to \(t=2\)

Step-by-Step Solution

Answer

The integral diverges to infinity.

Step 1: Understanding the Region of Integration

The region of integration is in the first quadrant and is bounded from below by the curve \(s = \ln t\) and from the left by \(t=1\). The region extends from \(t=1\) to \(t=2\). Thus, for a fixed \(t\), \(s\) ranges from \(\ln t\) to some upper limit. Since there is no explicit upper bound for \(s\), we conclude it extends upwards to infinity.

Step 2: Setting up the Integral

Based on the region of integration described, we set up the double integral. We first integrate with respect to \(s\) from \(s = \ln t\) to \(s = \infty\). Then, we integrate with respect to \(t\) from \(t = 1\) to \(t = 2\). This gives the integral \(\int_{1}^{2} \int_{\ln t}^{\infty} e^{s} \, ds \, dt.\)

Step 3: Solving the Inner Integral

The inner integral \(\int_{\ln t}^{\infty} e^{s} \, ds\) involves integrating \(e^{s}\) from \(s = \ln t\) to infinity. The antiderivative of \(e^{s}\) is \(e^{s}\), so we have\[\int_{\ln t}^{\infty} e^{s} \, ds = \lim_{A \to \infty} [e^{s}]_{\ln t}^{A} = \lim_{A \to \infty} (e^{A} - e^{\ln t}).\]This simplifies to \(\lim_{A \to \infty} e^{A} - t = \infty - t\).

Step 4: Evaluating the Outer Integral

The outer integral becomes \(\int_{1}^{2} (\infty - t) \, dt\). The term \(\infty - t\) clearly represents an improper integral that diverges to infinity, as \(\int_{1}^{2} \infty \, dt\) goes to infinity. Thus, no finite value can be assigned to this integral.

Step 5: Concluding the Integration

The integral diverges. Since one of the bounds of integration extends to infinity and involves an expression that grows without bound, the entire definite integral does not converge to a finite, real number.

Key Concepts

Improper Integrals

When dealing with improper integrals, it's crucial to recognize that one or more limits of integration are infinite or if the integrand exhibits discontinuities over the interval of integration. In our exercise, the inner integral is improper due to the upper limit of integration being infinity:
\[ \int_{\ln t}^{\infty} e^{s} \, ds. \]
Improper integrals are calculated by taking limits, allowing us to handle terms that involve infinity in a controlled way. For instance, when integrating the exponential function \(e^s\) from \(s = \ln t\) to infinity, we take the limit:
\[ \lim_{A \to \infty} [e^{s}]_{\ln t}^{A} = \lim_{A \to \infty} (e^{A} - e^{\ln t}) = \infty - t. \]
Recognizing the divergence, or non-convergence, is key when evaluating improper integrals. Here, the expression grows without bound, thus making the integral diverge and yield an infinite result.

• Improper integrals occur with infinite limits.
• Handling improper integrals often involves evaluating limits.
• Divergence indicates the integral does not settle to a finite answer.

Region of Integration

Understanding the region of integration provides clarity for setting up integrals properly. In this problem, the region is outlined in the first quadrant on the \(st\)-plane. It's given by:

• The region is bounded below by the curve \(s = \ln t\).
• The region begins at \(t = 1\) and stretches to \(t = 2\).
• For any selected \(t\), \(s\) starts at \(\ln t\) and extends upwards indefinitely.

This description guides the formation of a double integral by defining the limits within which each variable (\(s\) and \(t\)) is allowed to vary. Setting these limits accurately is essential to solving the integral correctly.
By visualizing the region as stretching infinitely upwards from \(s = \ln t\) over the segment \(t = 1\) to \(t = 2\), we can confidently establish that there’s no upper boundary for \(s\). This unbounded nature inherently brings improper integration into play.

Limits of Integration

The limits of integration are crucial as they define the interval over which the integration takes place. In this double integral exercise, the limits are based on the region of integration:

• The inner integral: \(s\) spans from \(s = \ln t\) to \(s = \infty\).
• The outer integral: \(t\) reaches from \(t = 1\) to \(t = 2\).

These limits must accurately reflect the domain of integration identified by the sketch or description of the region. For this case, because \(s\) extends to infinity, it introduces the aspect of improper integration, impacting the solution of the integral.
Correctly determining and using limits helps ensure that the calculations in solving the integral consider the designated area around which the function is being integrated. Bear in mind that one limit reaching infinity directly leads to an improper integral, presenting challenges such as potential divergence.