Problem 38 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{(\ln z)^{2}}{z} d z$$

Step-by-Step Solution

Answer

\( \int \frac{(\ln z)^2}{z} \, dz = \frac{(\ln z)^3}{3} + C \).

Step 1: Identify the Method for Integration

We have the integral \( \int \frac{(\ln z)^2}{z} \, dz \). Notice that this function seems like a candidate for a substitution method, specifically using substitution for \( \ln z \), since its derivative is \( \frac{1}{z} \).

Step 2: Choose the Substitution

Let \( u = \ln z \), which implies \( du = \frac{1}{z} \, dz \). Substituting into the integral, we get \( \int u^2 \, du \).

Step 3: Integrate with Respect to the New Variable

The integral \( \int u^2 \, du \) is a simple power integration. Using the power rule, we integrate to get \( \frac{u^3}{3} + C \), where \( C \) is the constant of integration.

Step 4: Substitute Back the Original Variable

Replace \( u \) with \( \ln z \) to express the integral in terms of \( z \). Thus, we have \( \frac{(\ln z)^3}{3} + C \) as the solution to the original integral.

Step 5: Verify by Differentiation

Differentiate \( \frac{(\ln z)^3}{3} + C \) with respect to \( z \). Applying the chain rule, the derivative is \( \frac{3(\ln z)^2}{3} \cdot \frac{1}{z} = \frac{(\ln z)^2}{z} \), which matches the original integrand, thus verifying the solution is correct.

Key Concepts

Method of Integration

The extit{Method of Integration}, particularly integration by substitution, is an essential concept in calculus that simplifies the process of finding integrals. Imagine you're facing a complex integral that seems difficult to tackle directly. By transforming it into a simpler form, you make it easier to evaluate.

• **Identifying Candidates:** Not every integral needs substitution. Look for functions that contain a composite function, where part of the function is the derivative of another part. This suggests using substitution.
• **Choosing the Right Substitution:** Once you've spotted a potential candidate, choose a substitution that simplifies the integral. For example, in our case, setting \( u = \ln z \) simplifies the integral drastically, because \( du = \frac{1}{z} dz \).
• **Transform and Integrate:** After choosing a substitution, rewrite the integral in terms of the new variable and integrate. This often turns a complicated integral into a straightforward one.
• **Don't Forget to Reverse:** Once integrated, substitute back to express the answer in terms of the original variable.

By following these simple steps, the method of substitution becomes a powerful tool in solving integrals.

Differentiation

Differentiation, the process of finding the derivative, is closely linked to integration. Remember, differentiation and integration are inverse processes.

• **The Role of Differentiation:** In the integration context, differentiation helps us verify our results. After finding an integral, differentiating the result should yield the original integrand if done correctly.
• **Easy Verification:** Take our solution, \( \int \frac{(\ln z)^2}{z} dz = \frac{(\ln z)^3}{3} + C \). To verify, differentiate it with respect to \( z \). Applying the chain rule, the derivative simplifies back to our original integrand, which confirms the correctness.
• **Chain Rule in Action:** Here, the chain rule is crucial, as it accounts for derivatives of composite functions. When differentiating \( \frac{(\ln z)^3}{3} \), you first apply the power rule, then multiply by the derivative of \( \ln z \). This process reflects the interconnected nature of differentiation and integration.

Using differentiation as a verification step ensures the integrity of your integral solutions and enhances your understanding of these two core calculus operations.

Power Rule

The extit{Power Rule} is a fundamental technique in calculus that simplifies both differentiation and integration.

• **For Integration:** When integrating \( x^n \), the power rule assists by stating that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \). It's direct and widely applicable for polynomials and functions that can be expressed in polynomial form.
• **Applying During Substitution:** In our problem, after substituting \( u = \ln z \), we integrated \( u^2 \, du \) using the power rule. This gave us \( \frac{u^3}{3} + C \), streamlining the process significantly.
• **Power Rule for Differentiation:** This rule also applies when differentiating, where \( \frac{d}{dx} x^n = nx^{n-1} \). Understanding this parallel helps in both analytic and numerical calculations.
• **A Simple Tool for Complex Problems:** Despite its simplicity, the power rule is instrumental in solving complex integrals after simplifying them via substitution. It underlies many steps in calculus and acts as a crucial baseline for students mastering integration and differentiation.

Mastering the power rule for both differentiation and integration opens up a vast array of calculus problems that can be efficiently tackled.