Problem 21 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int_{1}^{4} \ln \sqrt{x} d x $$

Step-by-Step Solution

Answer

\( \int_{1}^{4} \ln \sqrt{x} \, dx = 4 \ln 2 - \frac{3}{2} \).

Step 1: Identify Parts for Integration by Parts

The integration by parts formula is given by \( \int u \: dv = uv - \int v \: du \). We need to choose \( u \) and \( dv \). For the integral \( \int \ln \sqrt{x} \, dx \), we choose \( u = \ln \sqrt{x} \) and \( dv = dx \).

Step 2: Differentiate \( u \) and Integrate \( dv \)

Find \( du \) by differentiating \( u = \ln \sqrt{x} \). Recall that \( \ln \sqrt{x} = \frac{1}{2} \ln x \), so \( du = \frac{1}{2x} \, dx \). Integrate \( dv = dx \) to get \( v = x \).

Step 3: Apply 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 integral on the right to get \( x \ln \sqrt{x} - \frac{1}{2} \int 1 \, dx \).

Step 4: Simplify and Evaluate the Remaining Integral

The remaining integral is \( \int 1 \, dx = x \), so the expression becomes \( x \ln \sqrt{x} - \frac{1}{2}x \). Substitute the limits of integration \( 1 \) and \( 4 \) into the expression: Evaluate from \( 1 \) to \( 4 \): \( \left[ 4 \ln \sqrt{4} - \frac{1}{2}(4) \right] - \left[ 1 \ln \sqrt{1} - \frac{1}{2}(1) \right] \).

Step 5: Evaluate the Definite Integral

Evaluate \( \ln \sqrt{4} = \ln 2 \) and \( \ln \sqrt{1} = 0 \). Therefore, the expression becomes \( [4 \cdot \ln 2 - 2] - [0 - \frac{1}{2}] = 4 \ln 2 - 2 + \frac{1}{2} \), simplifying to \( 4 \ln 2 - \frac{3}{2} \).

Key Concepts

Techniques of Integration

Integration by parts is a crucial technique in calculus used to integrate products of functions. This method derives from the product rule for differentiation and is particularly useful when dealing with logarithmic, exponential, or polynomial functions.

For integration by parts, the formula \(\int u \: dv = uv - \int v \: du\) involves selecting which parts of the integrand to differentiate and which to integrate. Key steps include:

• Choosing \(u\) and \(dv\): Selecting the right \(u\) and \(dv\) is crucial. It's often best to let \(u\) be a function that simplifies upon differentiation, such as \( \ln(x) \).


• Differentiating \(u\) and integrating \(dv\): \(\frac{du}{dx}\) is found by differentiating \(u\), while \(v\) is the integral of \(dv\).


• Applying the Formula: Substitute \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula to find the integral.

Using integration by parts efficiently requires practice and sometimes creativity to choose the best functions for \(u\) and \(dv\). Always verify your results by differentiating back to check the solution.

Definite Integrals

Definite integrals calculate the net area under a curve between two points, providing valuable insight into the cumulative sum of values over an interval. When evaluating them, we adhere to these key principles:

• Compute the antiderivative of the function inside the integral. In the case of our problem, it was necessary to use the integration by parts method first, to find
  the integral before evaluating it between the limits 1 and 4.


• Apply the fundamental theorem of calculus, which states that for a function \( F \) that is continuous over an interval \([a, b]\), if \( f \) is an antiderivative
  of \( F \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

The definite integral takes into account areas above and below the x-axis, with areas below contributing negative values.
The careful evaluation of endpoints is crucial, as it determines the correct magnitude of the integral's result. Thus, knowing when and how to apply this fundamental theorem ensures accurate calculations.

Logarithmic Functions

Logarithmic functions, especially the natural logarithm \(\ln(x)\), frequently appear in calculus due to their unique properties. Logs transform complicated expressions into more manageable forms, which is particularly helpful for integration.

In this exercise, \(u\) is chosen as \( \ln \sqrt{x} \). Simplifying \( \ln \sqrt{x} \) to \( \frac{1}{2} \ln x \) was a key step. This simplification facilitated applying the integration by parts formula. Important points about logarithmic functions in integration include:

• Simplification: Recognizing transformations, like \( \ln(a^b) = b \ln(a) \), helps simplify integrals.


• Integration Techniques: Logarithmic properties aid in differentiating and integrating functions.
  Using properties like \( \ln x^n = n \ln x \) helps in breaking down complex expressions.

Logarithmic functions require a deep understanding to apply effectively within integration and will often lead to cleaner, simpler solutions when used correctly.