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Magazine Chapter 6: Problem 5
$$ \int_{1}^{4} \frac{\ln \sqrt{s}}{\sqrt{s}} d s $$
Short Answer
Expert verified
4 \(\text{ln}(4) - 6\)
Step by step solution
01
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: \(\frac{\frac{1}{2} \text{ln}(s)}{s^{1/2}} = \frac{1}{2} \frac{\text{ln}(s)}{s^{1/2}} = \frac{1}{2} s^{-1/2} \text{ln}(s)\).
02
Set Up the New Integral
Based on the simplification, the integral can be rewritten as: \(\frac{1}{2} \int_{1}^{4} s^{-1/2} \text{ln}(s) \, d s\).
03
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 \(v \).
04
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 \(v = 2s^{1/2}\).
05
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 \(2 \text{ln}(s) s^{1/2} - 2 \int s^{-1/2} \, d s\).
06
Integrate the Remaining Integral
Integrate \(2 \int s^{-1/2} \, d s\). This integral is straightforward and results in \(\int s^{-1/2} \, ds = 2s^{1/2}\).
07
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 \(\text{ln}(s) s^{1/2} - 2s^{1/2}\).
08
Evaluate at the Bounds
Evaluate the expression at the upper and lower bounds (4 and 1): \((\text{ln}(4) 4^{1/2} - 2 \cdot 4^{1/2}) - (\text{ln}(1) 1^{1/2} - 2 \cdot 1^{1/2})\).
09
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 \(4 \text{ln}(4) - 6\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \(\text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\)
- \(\text{ln}\left(\frac{a}{b}\right) = \text{ln}(a) - \text{ln}(b)\)
- \(\text{ln}(a^b) = b \text{ln}(a)\)
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:
- \(u = \text{ln}(s)\)
- \(dv = s^{-1/2} \, ds\)
