Problem 11 - Calculus and its applications

Question

Evaluate using integration by parts or substitution. (Assume \(u>0\) in \(\ln\) u. Check by differentiating. $$ \int x \ln \sqrt{x} d x $$

Step-by-Step Solution

Answer

Integral evaluated is \( \frac{x^2}{8} \ln x - \frac{x^2}{16} + C \).

Step 1: Rewrite the Integral

First, simplify the expression inside the integral: \( \ln \sqrt{x} = \frac{1}{2} \ln x \). Thus, the integral becomes: \[ \int x \cdot \frac{1}{2} \ln x \, dx = \frac{1}{2} \int x \ln x \, dx \]

Step 2: Integration by Parts Setup

Use the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Let \( u = \ln x \) and \( dv = x \, dx \). Then \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^2}{2} \).

Step 3: Apply Integration by Parts

Substitute into the integration by parts formula: \[ \frac{1}{2} ( \int x \ln x \, dx ) = \frac{1}{2} \left( \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \right) \] Simplify the inner integral: \( \int \frac{x^2}{2x} \, dx = \int \frac{x}{2} \, dx \).

Step 4: Solve the Remaining Integral

The integral simplifies to: \[ \frac{1}{2} \left( \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx \right) \]Calculate \( \int x \, dx = \frac{x^2}{2} \). Thus, the expression becomes:\[ \frac{1}{2} \left( \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} \right) \]

Step 5: Simplify the Expression

After simplifying, you get: \[ \frac{x^2}{4} \ln x - \frac{x^2}{8} \] Thus, the integral evaluates to: \( \frac{1}{2} \left( \frac{x^2}{4} \ln x - \frac{x^2}{8} \right) + C \), where \( C \) represents the constant of integration.

Step 6: Combine Terms

Combine and simplify the terms:\[ \frac{x^2}{8} \ln x - \frac{x^2}{16} + C \] This is the evaluated result of the integral.

Step 7: Verification by Differentiation

Differentiate \( \frac{x^2}{8} \ln x - \frac{x^2}{16} + C \). First term: Use the product rule on \( \frac{x^2}{8} \ln x \) to get: \( \frac{x}{4} \ln x + \frac{x}{8} \). Second term: Derivative of \( \frac{x^2}{16} \) is \( \frac{x}{8} \). Combine to verify: \( \frac{x}{8} \ln x \), which matches the integrand \( \frac{x}{2} \ln x \), validating our solution.

Key Concepts

Definite Integration

Definite integration is about finding the net area under a curve within a given interval. It effectively sums up infinite small parts under the curve between two points on the x-axis. These points are called the upper and lower limits of integration. To apply definite integration, you set up the integral with its limits. For example, if we look at the expression \( \int_{a}^{b} f(x) \, dx \), \( a \) and \( b \) are the limits of integration. The function \( f(x) \) is integrated across these limits. Once integrated, the fundamental theorem of calculus tells us that we can evaluate the integral at these points and subtract the results to find the net area:

• If \( F(x) \) is the integral of \( f(x) \), then \( F(b) - F(a) \) gives us the definite integral.

For our example in indefinite integration, this process isn't used because no limits of integration are given. However, understanding this process is crucial for problems where limits are involved, as you'll often simplify the integral first before applying these steps.

Logarithmic Functions

Logarithmic functions, notably the natural logarithm \( \ln x \), are central in a variety of mathematical contexts. They are the inverse of exponential functions and have unique properties that simplify integration and differentiation. For instance, the integral of \( \ln x \) is a classic example where integration by parts is effectively used, as seen in our exercise. When you simplify \( \ln \sqrt{x} \) to \( \frac{1}{2} \ln x \), it demonstrates one useful property: \( \ln a^b = b \ln a \). Here, \( \ln \sqrt{x} \) becomes \( \frac{1}{2} \ln x \) because \( \sqrt{x} \) is \( x^{1/2} \). Logarithmic functions have distinct characteristics:

• \( \ln 1 = 0 \)
• The natural log of a product: \( \ln(xy) = \ln x + \ln y \)
• Base change formula: \( \log_b x = \frac{\ln x}{\ln b} \), useful in computations involving different bases.

These features make logarithmic functions versatile tools in solving complex integration problems.

Differentiation Verification

Differentiation verification is the process of confirming an integration by checking the derived function through differentiation. This step ensures that the antiderivative is correct since differentiating should return the original integrand. In our exercise, after integrating, we get the function \( \frac{x^2}{8} \ln x - \frac{x^2}{16} + C \). Differentiating involves the product rule and fundamental differentiation skills:

• The derivative of \( \frac{x^2}{8} \ln x \) requires the product rule, which states \( (uv)' = u'v + uv' \).
• Applying to \( \frac{x^2}{8} \ln x \), where \( u = \frac{x^2}{8} \) and \( v = \ln x \), yields \( \frac{x}{4} \ln x + \frac{x}{8} \).
• The derivative of \( -\frac{x^2}{16} \) simplifies to \( -\frac{x}{8} \).
• Adding these derivatives gives \( \frac{x}{8} \ln x \), matching \( \frac{x}{2} \ln x \), thereby confirming the integration.

This reassurance through differentiation is a critical step in problem solving, guaranteeing that the integral was done correctly.