Problem 21 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{1}^{4} \ln \sqrt{x} d x $$

Step-by-Step Solution

Answer

The value of the integral is 2.5.

Step 1: Understanding Integration by Parts Formula

Integration by parts is based on the product rule for differentiation and is given by the formula:\[\int u \, dv = uv - \int v \, du\]In this problem, we need to appropriately select \(u\) and \(dv\) from the integral \(\int \ln \sqrt{x} \, dx\) for further simplification.

Step 2: Selecting u and dv

We select \(u = \ln \sqrt{x}\) and \(dv = dx\). This is a strategic choice, as differentiating \(\ln \sqrt{x}\) simplifies the expression. Hence, we have:\[du = \frac{1}{2x} dx\]For \(dv = dx\), we integrate to find \(v = x\).

Step 3: Applying Integration by Parts Formula

Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula:\[\int \ln \sqrt{x} \, dx = x \ln \sqrt{x} - \int x \cdot \frac{1}{2x} \, dx\]Simplify the expression inside the integral:\[\int \frac{1}{2} \, dx\]

Step 4: Simplifying and Evaluating the Integral

Compute the remaining integral:\[\int \frac{1}{2} \, dx = \frac{1}{2}x + C\]Thus, the integral of the original function is:\[\int \ln \sqrt{x} \, dx = x \ln \sqrt{x} - \frac{1}{2}x + C\]

Step 5: Calculating Definitive Integral

Substitute the bounds \([1, 4]\) into the expression:\[\left[ x \ln \sqrt{x} - \frac{1}{2}x \right]_1^4 = \left( 4 \ln \sqrt{4} - \frac{1}{2}(4) \right) - \left( 1 \ln \sqrt{1} - \frac{1}{2}(1) \right)\]This simplifies to:\[\left( 4 \cdot 1 - 2 \right) - (0 - 0.5) = 2 + 0.5 = 2.5\]

Key Concepts

Understanding Definite Integrals

Definite integrals allow us to calculate the accumulated value over a specific interval. In this example, the bounds are from 1 to 4. This means we are determining the total area under the curve of the function \(\ln \sqrt{x}\) on this interval. The key aspect of definite integrals is their fixed bounds:

• Lower Bound (1): The starting point of accumulation.
• Upper Bound (4): The ending point of accumulation.

When solving for definite integrals, the evaluation process requires substituting the upper and lower bounds into the antiderivative. Then, subtract the antiderivative at the lower bound from that at the upper bound. This provides the net area:

• The antiderivative serves as the basis for substitution.
• Use the Fundamental Theorem of Calculus to transition from antiderivative to definite integral calculation.

For our problem, this process follows through carefully, eventually simplifying to a numeric solution of 2.5, providing clear insight into the integral's result over the given interval.

Principles of Logarithmic Integration

Logarithmic integration is crucial when dealing with integrals that include logarithms. Such integrals often require strategic approaches due to the natural logarithm's unique properties. In our example, we work with \(\ln \sqrt{x}\).Let's break down the core elements of logarithmic integration:

• Logarithmic Derivative Key Insight: The derivative of \(\ln x\) is \(\frac{1}{x}\), often simplifying integral expressions.
• Simplified Expression: Reducing this example to \(\ln \sqrt{x} = \frac{1}{2} \ln x\) simplifies derivative evaluations.

In the integration by parts approach, understanding these logarithmic properties helps in selecting the correct parts (like \(u\) and \(dv\)). This strategic selection, when mixed with prior simplifications, refines the integration path. The derivative of the chosen \(u\) function, \(({\ln \sqrt{x}})\) minimizes complexity, letting us solve easily.

Effective Calculus Problem Solving Techniques

With calculus problem solving, focus on identifying optimal strategies according to the integral's structure. The integration by parts technique requires:

• Selection Strategy: Choosing the best \(u\) and \(dv\) simplifies derivatives and integrals subsequently.
• Integration by Parts Formula: Recognize the repeated use of \(\int u \, dv = uv - \int v \, du\) across problems, but customized for specific functions.

Steps to solve calculus problems efficiently include:

Systematic Analysis

Identify components like logarithms or polynomials that dictate the best method (such as parts, substitution).

Execution and Simplification

Calculate derivatives and integrals carefully, reducing at each stage to prevent errors.By tackling problems systematically, like setting bounds appropriately and simplifying expressions, success in calculus problems becomes more attainable. Reflect on the task at hand, and apply an optimum strategy fitting the specific characteristics of the integral.