Problem 63 - Calculus Single Variable

Question

$$ \text { Evaluate } \int x^{p} \ln (x) d x $$

Step-by-Step Solution

Answer

\( \int x^p \ln(x) \, dx = \frac{x^{p+1} \ln(x)}{p+1} - \frac{x^{p+1}}{(p+1)^2} + C \).

Step 1: Select a method

To solve the integral \( \int x^p \ln(x) \, dx \), we should use integration by parts. This method is suitable because the integrand is a product of two functions: \( x^p \) and \( \ln(x) \).

Step 2: Identify parts for integration by parts

Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). We choose \( u = \ln(x) \), which makes \( du = \frac{1}{x} \, dx \). Let \( dv = x^p \, dx \), so \( v = \frac{x^{p+1}}{p+1} \).

Step 3: Apply the integration by parts formula

Substituting into the integration by parts formula, we have \( \int x^p \ln(x) \, dx = \ln(x) \cdot \frac{x^{p+1}}{p+1} - \int \frac{x^{p+1}}{p+1} \, \frac{1}{x} \, dx \).

Step 4: Simplify the new integral

The new integral becomes \( \int \frac{x^{p+1}}{p+1} \, \frac{1}{x} \, dx = \int \frac{x^p}{p+1} \, dx \). This simplifies to \( \frac{1}{p+1} \int x^p \, dx \).

Step 5: Evaluate the remaining integral

The integral \( \int x^p \, dx \) is \( \frac{x^{p+1}}{p+1} \) (for \( p eq -1 \)). Therefore, \( \frac{1}{p+1} \int x^p \, dx = \frac{x^{p+1}}{(p+1)^2} \).

Step 6: Substitute back into the integration by parts formula

The complete solution is \( \int x^p \ln(x) \, dx = \frac{x^{p+1} \ln(x)}{p+1} - \frac{x^{p+1}}{(p+1)^2} + C \), where \( C \) is the constant of integration.

Key Concepts

Calculus Integration Techniques

When it comes to integrating functions, choosing the right technique is essential to simplify the problem. In this exercise, we are focusing on integration by parts, a powerful method used when integrating the product of two functions. It is particularly handy in situations where separating two complex functions can reveal simpler parts to integrate.

According to the formula for integration by parts: \[\int u \, dv = uv - \int v \, du,\]we first identify the components within our integral that can be labeled as \( u \) and \( dv \). A strategic choice here can streamline the process significantly.

• We usually pick \( u \) to be the function whose derivative, \( du \), simplifies the integral.
• The choice of \( dv \) should be a function that, when integrated to become \( v \), does not complicate the integral further.

By choosing a suitable \( u \) and \( dv \), the integration reduces to evaluating simpler integrals, thus making calculations more efficient.

Definite and Indefinite Integrals

Integrals are fundamental in calculus and come in two forms: definite and indefinite. Understanding their differences and applications is key to mastering calculus.

Indefinite Integrals capture the family of all antiderivatives of a function. They do not have bounds and include an arbitrary constant, \( C \), because adding a constant to a function's antiderivative does not affect its derivative. Our exercise provided an indefinite integral, as shown with the presence of the \( C \) in the final solution.

• The indefinite integral of a function \( f(x) \) is generally denoted by \( \int f(x)\,dx = F(x) + C \).

Definite Integrals consider limits of integration, thus providing numerical values. These integrals compute the net area between the function and the x-axis over a specific interval. They transform a function into a number, representing the signed area.

• Definite integrals are expressed as \( \int_{a}^{b} f(x)\,dx \), indicating the evaluation between limits \( a \) and \( b \).

Mastering both forms is crucial for solving a range of problems in calculus, from basic to complex applications.

Logarithmic Functions

Logarithmic functions, such as \( \ln(x) \), have distinctive characteristics that make them central in calculus. Their appearance within integrals can complicate processes but are manageable with techniques such as integration by parts.

The natural logarithm \( \ln(x) \) functions as an inverse to exponentials, which is reflected through its unique properties:

• \( \ln(e) = 1 \) since \( e^1 = e \).
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \).
• The integral of \( \ln(x) \) by standard methods isn't straightforward and often requires integration by parts.

In the given exercise, we set \( u = \ln(x) \) because its derivative \( \frac{1}{x} \) aligns nicely with simplifying the integral through integration techniques.
Understanding these properties helps explain why logarithms, though sometimes tricky, are fundamentally solvable within integrals.

Polynomial Functions

Polynomial functions are incredibly versatile and commonly appear in calculus problems due to their straightforward derivatives and integrals.

A polynomial with degree \( p \), such as \( x^p \), simplifies integration, which is inherently advantageous, especially when evaluating indefinite integrals:

• The integral of \( x^p \), given \( p eq -1 \), is \( \frac{x^{p+1}}{p+1} \)+C.
• This formula easily extends to polynomials of higher degrees.

In the solution, \( x^p \) takes the role of \( dv \) in the integration by parts because integrating \( x^p \) is direct and feasible.Thus, leveraging the simplicity of polynomial integration facilitates handling more complex integral problems by breaking them down into manageable pieces.