Problem 5 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

$$ \int_{1}^{4} \frac{\ln \sqrt{s}}{\sqrt{s}} d s $$

Step-by-Step Solution

Answer

4 \(\text{ln}(4) - 6\)

Step 1: Simplify the Integrand

Rewrite the integrand \(\frac{\text{ln}\big(\text{sqrt}(s)\big)}{\text{sqrt}(s)}\) using logarithmic properties and simplifying. We know that \(\text{ln}(\text{sqrt}(s)) = \frac{1}{2} \text{ln}(s)\) and \(\text{sqrt}(s) = s^{1/2} \). Therefore, the integrand simplifies to: \(\frac{\frac{1}{2} \text{ln}(s)}{s^{1/2}} = \frac{1}{2} \frac{\text{ln}(s)}{s^{1/2}} = \frac{1}{2} s^{-1/2} \text{ln}(s)\).

Step 2: Set Up the New Integral

Based on the simplification, the integral can be rewritten as: \(\frac{1}{2} \int_{1}^{4} s^{-1/2} \text{ln}(s) \, d s\).

Step 3: Use Integration by Parts

To integrate \(\frac{1}{2} \int_{1}^{4} s^{-1/2} \text{ln}(s) \, d s\), use the integration by parts formula, \(\int u \, dv = uv - \int v \, du\). Let \(u = \text{ln}(s)\) and \(dv = s^{-1/2} \, d s\). Then, differentiate and integrate to find \(du \) and \(v \).

Step 4: Calculate du and v

Differentiate \(u = \text{ln}(s)\) to find \(du = \frac{1}{s} \, d s\). Integrate \(dv = s^{-1/2} \, d s\) to find \(v = 2s^{1/2}\).

Step 5: Apply Integration by Parts Formula

Now apply the integration by parts formula: \(\int \text{ln}(s) s^{-1/2} \, d s = \text{ln}(s) \, 2s^{1/2} - \int 2s^{1/2} \frac{1}{s} \, d s\) which simplifies to \(2 \text{ln}(s) s^{1/2} - 2 \int s^{-1/2} \, d s\).

Step 6: Integrate the Remaining Integral

Integrate \(2 \int s^{-1/2} \, d s\). This integral is straightforward and results in \(\int s^{-1/2} \, ds = 2s^{1/2}\).

Step 7: Combine and Simplify

Combine the results: \(\frac{1}{2} [2 \text{ln}(s) s^{1/2} - 4s^{1/2}]\), and factor out common terms. The integral simplifies to \(\text{ln}(s) s^{1/2} - 2s^{1/2}\).

Step 8: Evaluate at the Bounds

Evaluate the expression at the upper and lower bounds (4 and 1): \((\text{ln}(4) 4^{1/2} - 2 \cdot 4^{1/2}) - (\text{ln}(1) 1^{1/2} - 2 \cdot 1^{1/2})\).

Step 9: Simplify the Evaluated Expression

Since \(\text{ln}(1) = 0\), the expression simplifies to \((2 \cdot 2 \text{ln}(4) - 4) - (0 - 2) \). This simplifies further to \(4 \text{ln}(4) - 6\).

Key Concepts

Definite Integral

The concept of a definite integral is crucial in calculus. It allows us to calculate the exact area under a curve within a specific interval. For instance, in our problem, we aim to determine the area under the curve of \(\frac{\text{ln}(\text{sqrt}(s))}{\text{sqrt}(s)}\) from s = 1 to s = 4. The definite integral is written as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the lower and upper bounds, respectively.

Logarithmic Properties

Logarithmic properties are incredibly useful for simplifying integrals. One such property is \(\text{ln}(\text{sqrt}(s)) = \frac{1}{2} \text{ln}(s)\), which we used in our problem to rewrite the integrand. Other key properties include:

• \(\text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\)
• \(\text{ln}\left(\frac{a}{b}\right) = \text{ln}(a) - \text{ln}(b)\)
• \(\text{ln}(a^b) = b \text{ln}(a)\)

These properties can simplify complex logarithmic expressions, making integration easier.

Calculus

Calculus encompasses a variety of methods and principles to explore change and areas. One important technique is integration by parts, used when integrating products of functions. The formula is \( \int u \, dv = uv - \int v \, du\). In our problem, we set:

• \(u = \text{ln}(s)\)
• \(dv = s^{-1/2} \, ds\)

This allowed us to compute the definite integral effectively. Calculus also involves understanding limits, derivatives, and the Fundamental Theorem of Calculus, which links differentiation and integration.