Problem 22 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The integral evaluates to approximately 1.861.

Step 1: Identify Parts for Integration by Parts

For integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). In this problem, we set \( u = \ln \sqrt{x} \) which simplifies to \( u = \frac{1}{2}\ln{x} \) and \( dv = \sqrt{x} \, dx \), making \( dv = x^{1/2}\, dx \).

Step 2: Differentiate and Integrate Parts

Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \):\[ du = \frac{1}{2x} \, dx \] \[ v = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} \] Thus, the differential and integral parts are \( du = \frac{1}{2x} \, dx \) and \( v = \frac{2}{3} x^{3/2} \).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:\[ \int \sqrt{x} \ln \sqrt{x} \, dx = \left[ \frac{1}{2} \ln{x} \cdot \frac{2}{3} x^{3/2} \right]_1^4 - \int_1^4 \frac{2}{3} x^{3/2} \cdot \frac{1}{2x} \, dx \].

Step 4: Simplify the Resulting Expression

Calculate the first term by evaluating \( \left[ \frac{1}{2} \ln{x} \cdot \frac{2}{3} x^{3/2} \right] \) from 1 to 4 and simplify the second linked integral:\[ = \left[ \frac{2}{6} x^{3/2} \ln{x} \right]_1^4 - \int_1^4 \frac{1}{3} x^{1/2} \, dx \].

Step 5: Evaluate the Remaining Integral

Evaluate the integral \( \int \frac{1}{3} x^{1/2} \, dx \):\[ = \frac{1}{3} \cdot \left[ x^{3/2} \cdot \left( \frac{2}{3} \right) \right]_1^4\]\[ = \frac{2}{9} \cdot \left[ x^{3/2} \right]_1^4 \].

Step 6: Calculate Definite Integrals

Calculate the definite integrals separately:1. Evaluate \( \left[ \frac{2}{6} x^{3/2} \ln{x} \right]_1^4 \) giving: \[ \frac{2}{6} \left( (4^{3/2} \ln{4}) - (1^{3/2} \ln{1}) \right) = \frac{8}{3} \ln{4} \] 2. Evaluate \( \frac{2}{9} \left[ x^{3/2} \right]_1^4 \) giving: \[ \frac{2}{9} \left( 4^{3/2} - 1^{3/2} \right) = \frac{10}{3} \].

Step 7: Calculate Final Answer

Subtract results from Steps 6.1 and 6.2 to find the final answer:\[ \int_{1}^{4} \sqrt{x} \ln \sqrt{x} \, dx = \frac{8}{3} \ln{4} - \frac{10}{3} \] which simplifies to approximately \( 1.861 \) when computed numerically.

Key Concepts

Definite Integral

A definite integral is more than just finding an antiderivative. It calculates the accumulation of quantities, giving a real-world meaning to the area under a curve between two points on a graph. In our example, the definite integral is evaluated from 1 to 4. This means we are specifically interested in the area between the curve, the x-axis, and the vertical lines at x = 1 and x = 4.
When handling definite integrals, it's essential to understand its notation: \( \int_{a}^{b} f(x) \, dx \). Here, \( f(x) \) is the function being integrated, and [a, b] represents the interval over which we find the area.
Now, compute the definite integral by first finding the antiderivative and then applying the Fundamental Theorem of Calculus. This theorem states that if F is the antiderivative of f on an interval [a, b], then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

In the problem above, after using integration by parts, we compute two integrals separately over the interval [1, 4]. Each of these pieces contributes to finding the final answer by evaluating them at the boundaries and subtracting the results as shown in the solution.

Logarithmic Function

The logarithmic function, often denoted as \( \ln x \), is fundamental in integration, particularly when combined with power functions. Logarithms transform multiplicative relationships into additive ones, which is quite useful in integration by parts.
In the problem, we use \( \ln \sqrt{x} \) which simplifies to \( \frac{1}{2} \ln x \). This transformation is due to the property \( \ln a^b = b \ln a \). Understanding this property helps simplify expressions and makes differentiation straightforward.
Logarithmic functions are known for their slow growth. This means that they increase at a much slower rate compared to linear or power functions. This characteristic is important not just in calculus for integration purposes, but also in solving real-world problems like understanding growth trends or scaling issues.
When differentiating \( \ln x \), the derivative is \( \frac{1}{x} \), which changes the problem dynamics in integration by introducing a fraction. This interesting mix of functions can lead to more complex but solvable integrals, as you’ve seen in the problem solved through integration by parts.

Power Function

Power functions, such as \( x^{n} \), are some of the most fundamental in mathematics. In integrals, they show up often due to their simplicity and inherent properties. The problem involves \( \sqrt{x} \), which can be rewritten as \( x^{1/2} \).
The magic of power functions lies in their straightforward integration and differentiation rules. Specifically, the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), assuming \( n eq -1 \). This rule is essential as it simplifies many problems. For instance, integrating \( x^{1/2} \) gives \( \frac{2}{3} x^{3/2} \), as seen in the problem's integration steps.
Power functions are prevalent in nature and physics. From calculating area, volume, and motion, they model countless phenomena. Their predictable behavior and simple form make them perfect candidates for such models.
In solving complex integrals, power functions combined with logarithms, as you experienced, are a frequent match due to their complementary properties, adding layers to the problem but offering a path to an elegant solution through methods like integration by parts.