Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

apply integration by parts twice to evaluate each integral. $$ \int \ln ^{2} x^{20} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{20} [(400\ln^2 x - 40\ln x + 40)x] + C \).

Step 1: Apply Substitution

First, let's simplify the integral by substituting \( u = \ln(x^{20}) \), which is equivalent to \( u = 20 \ln x \). This simplifies our integrand but we need to adjust the differential as well. By differentiating, we get \( du = \frac{20}{x} \, dx \). Thus, \( dx = \frac{x}{20} \, du \).

Step 2: Rewrite the Integral

Rewrite the integral in terms of \( u \):\[ \int (20\ln x)^2 \frac{x}{20} \, du = \frac{1}{20} \int u^2 x \, du \].

Step 3: Expression for x in terms of u

From our substitution \( u = 20 \ln x \), express \( x \) in terms of \( u \): \( x = e^{u/20} \).

Step 4: Integral in Terms of u

Substitute \( x = e^{u/20} \) into the integral:\[ \frac{1}{20} \int u^2 e^{u/20} \, du \].

Step 5: First Integration by Parts Setup

To apply integration by parts, choose\[ v = e^{u/20}, \ dv = \frac{1}{20}e^{u/20} \, du, \ u = u^2, \ du = 2u \, du \].

Step 6: First Integration by Parts Application

Using integration by parts \( \int u \, dv = uv - \int v \, du \):\[ uv = u^2 e^{u/20} \]Substituting and rearranging gives:\[ \int e^{u/20} u^2 \, du = u^2 e^{u/20} - \int e^{u/20} 2u \, du \].

Step 7: Second Integration by Parts Setup

We need to resolve \( \int e^{u/20} 2u \, du \) by integration by parts again:Choosing parts as:\[ v = e^{u/20}, \ dv = \frac{1}{20}e^{u/20} \, du, \ u = 2u, \ du = 2 \, du \].

Step 8: Second Integration by Parts Application

Applying integration by parts:\[ \int 2u e^{u/20} \, du = 2u e^{u/20} - \int 2 e^{u/20} \, du = 2u e^{u/20} - 40 e^{u/20} \].

Step 9: Substitute Back into Original Integral

Now substitute back into the formula:\[ \int u^2 e^{u/20} \, du = u^2 e^{u/20} - (2u e^{u/20} - 40 e^{u/20}) \].

Step 10: Simplifying the Expression

Simplify:\[ \frac{1}{20} (u^2 e^{u/20} - 2u e^{u/20} + 40 e^{u/20}) = \frac{1}{20} e^{u/20} (u^2 - 2u + 40) \].

Step 11: Reverse the Substitution

Reverse the substitution: Recall that \( u = 20 \ln x \), hence we have:\[ = \frac{1}{20} x^{u/20} (u^2 - 2u + 40) \].

Step 12: Final Expression in Terms of x

Rewrite the final integral back in terms of \( x \): \[ \frac{1}{20} (20\ln x)^2 x - 2 (20\ln x) x + 40x + C \].Thus,\[= \frac{1}{20} [(400 \ln^2 x - 40 \ln x + 40)x] + C\].

Key Concepts

Substitution Method

The substitution method, often referred to as the "u-substitution," is a strategic way to simplify integrals. It involves changing the variable of integration to make the expression easier to integrate. In the given problem, we started by letting \( u = 20 \ln x \). This step is crucial because it transforms the complex function \( \ln^2 x^{20} \) into a simpler form.

• Start by identifying a part of the integral that can easily be replaced with a single variable \( u \).
• In this problem, \( u = 20 \ln x \) was chosen to simplify the expression \( \ln^2 x^{20} \).
• Differentiate \( u = 20 \ln x \) to find \( du = \frac{20}{x} \, dx \), allowing us to replace \( dx \) in the original integral.
• Rewrite the entire integral in terms of \( u \), reducing complex operations to basic polynomial or exponential operations.

This change of variables simplifies the integral and sets it up perfectly for integration by parts.

Differential Calculus

Differential calculus plays an important role in the substitution method and integration by parts. Understanding how to differentiate various functions is essential because it allows us to replace differentials in integrals. In this exercise, following the substitution step, we used differentiation to express \( du \) in terms of \( dx \).

• Differential calculus involves finding derivatives and understanding the rate at which functions change.
• By differentiating \( u = 20 \ln x \), we obtained \( du = \frac{20}{x} \, dx \), which transforms how we integrate with respect to \( u \) instead of \( x \).
• This process showcases how calculus tools are employed to manipulate integrals, allowing a substitution that makes complex integrals manageable.

These concepts highlight the essential role of differentials in translating between variables during integration.

Transcendental Functions

Transcendental functions, such as logarithmic and exponential functions, are central to calculus. In this exercise, both the natural logarithm and the exponential function are key players in solving the integral.

• A transcendental function is not algebraic and includes functions like \( \ln(x) \) and \( e^x \).
• In the substitution step, the integral involves \( \ln^2 x^{20} \), which requires manipulation using the properties of logarithms.
• Later, when expressing the integral in terms of \( u \), the exponential function \( e^{u/20} \) appears, showcasing the interdependency of these functions in solving complex integrals.
• Understanding how to handle these transcendental functions is crucial for converting integrals into forms that are easier to evaluate analytically.

Mastery of such functions is indispensable when dealing with integration techniques, such as parts and substitution, involving transcendental expressions.