Problem 342 - Calculus Volume 1

Question

In the following exercises, verify by differentiation that \(\int \ln x d x=x(\ln x-1)+C, \quad\) then use appropriate changes of variables to compute the integral. $$\int \ln x d x \text { (Hint: } \int \ln x d x=\frac{1}{2} \int x \ln \left(x^{2}\right) d x )$$

Step-by-Step Solution

Answer

The integral \( \int \ln x \, dx \) is verified as \( x(\ln x - 1) + C \).

Step 1: Verify the Given Integral by Differentiation

To verify the integral \( \int \ln x \, dx = x(\ln x - 1) + C \), we differentiate the function \( x(\ln x - 1) + C \) with respect to \( x \).We have:\[ \frac{d}{dx}(x(\ln x - 1) + C) = \frac{d}{dx}(x \ln x - x) \]Apply the product rule to \( x \ln x \):\[ \frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]Thus,\[ \frac{d}{dx}(x \ln x - x) = (\ln x + 1) - 1 = \ln x \]Since the derivative simplifies to \( \ln x \), the given expression is verified as the integral.

Step 2: Use Substitution to Compute the Given Integral

We are given the hint \( \int \ln x \, dx = \frac{1}{2} \int x \ln(x^2) \, dx \). Let's calculate this using the substitution provided.Consider:\[ \int x \ln(x^2) \led dx \]Use substitution, let \( u = x^2 \). Then, \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} \ du \).The integral becomes:\[ \frac{1}{2} \int \ln(u) \, du \]Evaluate this integral:Let \( v = \ln u \), then \( \frac{dv}{du} = \frac{1}{u} \) and therefore, the integral of \( \ln u \) is given by "Integration by Parts" as:\[ \int \ln u \, du = u(\ln u - 1) + C \]Now, back-substitute \( u = x^2 \):\[ \frac{1}{2} [x^2 (\ln(x^2) - 1)] + C = \frac{1}{2} [x^2 (2\ln x - 1)] + C = x^2 (\ln x - \frac{1}{2}) + C \]Simplifying gives:\[ x(\ln x - 1) + C \] which is consistent with the given integral.

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus, particularly in differentiation. It is used when you need to differentiate a function that is the product of two other functions. If you have functions \(u(x)\) and \(v(x)\), their product \(u(x) \cdot v(x)\) can be differentiated as follows:The product rule formula says:

• \( \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

In simple terms, you differentiate the first function, multiply it by the second function, and then add it to the product of the first function and the derivative of the second.In our exercise, the product rule was used to differentiate \( x\ln x \). This involved treating \( x \) as one function and \( \ln x \) as another. Applying the formula, we derived the sum of \( \ln x \) and \( 1 \), ensuring the effective verification of the integral \( \int \ln x \, dx \).

Substitution Method

The substitution method is a powerful technique used in integration to simplify complex expressions. It often involves a change of variables to transform the original integral into a simpler form.The substitution method follows these general steps:

• Choose a substitution that simplifies the integral—usually a part of the integrand.
• Find the differential of the new variable.
• Substitute both the variable and the differential into the integral.

In the given exercise, we used substitution by letting \( u = x^2 \). This choice redefined the integral \( \int x \ln(x^2) \, dx \) into a more manageable form \( \frac{1}{2} \int \ln(u) \, du \). After calculating this, we back-substituted to return to the original variable, thus simplifying the entire integration process.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. It measures how a function changes as its input changes. In calculus, differentiation forms the foundation of concepts like velocity and slope.Key rules that aid in differentiation include:

• The product rule, which we’ve previously explained.
• The sum rule, allowing term-by-term differentiation.

In our exercise, differentiation was crucial in verifying the integral \( \int \ln x \, dx \). We started by differentiating the expression \( x(\ln x - 1) + C \). Each component of the function was differentiated using basic rules, ensuring that the outcome matched the original integrand, \( \ln x \). This verification step confirmed our initial integral solution.

Natural Logarithm

The natural logarithm \( \ln x \) is the logarithm to the base \( e \), where \( e \approx 2.718 \). It is a significant function in calculus, especially in integration and differentiation.Important properties of natural logarithms include:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln(a^n) = n \ln a \)
• The derivative of \( \ln x \) is \( \frac{1}{x} \)

In the context of the given problem, \( \ln x \) was the primary function involved in the integration. Understanding its properties is vital for applying methods like integration by parts or substitution effectively. Working through the integration of \( \ln x \), we encountered its straightforward derivative, verifying the correctness of the integral \( \int \ln x \, dx = x(\ln x - 1) + C \).