Problem 31 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int x^{n} \ln a x d x \quad(a \neq 0, n \neq-1)\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Parts

For integration by parts, we need to identify functions to assign as \(u\) and \(dv\) from the integral \(\int x^n \ln(ax) \, dx\). Choose \(u = \ln(ax)\) because its derivative simplifies nicely and \(dv = x^n \, dx\).

Step 2: Find Derivatives and Integrals

Calculate \(du\) and \(v\). Differentiating \(u = \ln(ax)\), we get \(du = \frac{1}{ax}a \, dx = \frac{1}{x} \, dx\). Integrate \(dv = x^n \, dx\) to find \(v\). Thus, \(v = \frac{x^{n+1}}{n+1}\), assuming \(n eq -1\).

Step 3: Apply Integration by Parts Formula

The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\). Substituting our variables, we have: \(\int x^n \ln(ax) \, dx = \ln(ax) \cdot \frac{x^{n+1}}{n+1} - \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx\).

Step 4: Simplify and Integrate

Simplify \(\int v \, du\): \(\int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx = \int \frac{x^n}{n+1} \, dx\). This integral is \(\frac{x^{n+1}}{(n+1)^2}\) after integrating. Substitute back to get the solution.

Step 5: Write the Final Answer

Combine all parts to write the complete solution: \(\int x^n \ln(ax) \, dx = \frac{x^{n+1}}{n+1} \ln(ax) - \frac{x^{n+1}}{(n+1)^2} + C\), where \(C\) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals are a key concept in integral calculus that concern finding the area under a curve between two specific boundaries on the x-axis. They are noted by the integral sign with limits of integration, which generally look like this: \\[ \int_{a}^{b} f(x) \, dx \] \where \(a\) and \(b\) are the lower and upper limits, respectively. Unlike indefinite integrals, which include a constant of integration \(C\), definite integrals yield a specific numeric value.

The process involves first finding the antiderivative of the function, and then calculating the difference between its values at the upper and lower limits. This calculation effectively gives the "net area", where areas above the x-axis are positive and below are negative. When using the Fundamental Theorem of Calculus, we express this as: \\[ F(b) - F(a) \] \where \(F(x)\) is the antiderivative of \(f(x)\). This result is crucial in many applications such as physics, economics, and engineering, where the concepts of net change and accumulation are significant.

Here is a quick summary of its properties:

• Linearity: \( \int (cf(x) + dg(x)) \, dx = c \int f(x) \, dx + d \int g(x) \, dx \).
• If \(f(x)\) is continuous on \([a, b]\), then \(\int_{a}^{b} f(x) \, dx\) exists.
• Reversing limits changes the sign: \(\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \).
• Adding integrals: \(\int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx = \int_{a}^{c} f(x) \, dx \).

Integral Calculus

Integral calculus is the branch of mathematics dealing with integrals and their applications. It is central to solving a variety of problems in science and engineering, with key tasks being the calculation of areas, volumes, central points, and other related quantities. The core operations include finding antiderivatives and evaluating definite or indefinite integrals.

When we talk about integration in calculus, there are two types:

• **Indefinite Integrals**: Represent a family of functions and include a constant of integration \(C\). It is represented as: \\[ \int f(x) \, dx = F(x) + C \]
• **Definite Integrals**: Provide the "net area" under the curve over a specified interval and don't involve \(C\): \\[ \int_{a}^{b} f(x) \, dx \]

Integration by parts, a technique derived from the product rule of differentiation, further helps solve integrals where functions are multiply intertwined—like the problem example \\( \int x^n \ln(ax) \, dx \).

Functions are chosen for \(u\) and \(dv\) based on how easily they can be differentiated and integrated. The formula \\[ \int u \, dv = uv - \int v \, du \] \allows us to replace a complicated integral with one that is easier to solve. This is used quite often when dealing with products of polynomial and logarithmic or exponential functions.

Logarithmic Functions

Logarithmic functions are functions involving logarithms, typically characterized by the presence of \(\ln x\) (natural logarithm) or \(\log_b x\) (logarithm base \(b\)). They are the inverse of exponential functions and play an essential role in both differential and integral calculus.

Key properties of logarithmic functions include:

• \(\ln(ab) = \ln a + \ln b\)
• \(\ln(a^b) = b \ln a\)
• \(\ln(1) = 0\)
• \(\ln(e) = 1\), where \(e\) is the natural logarithm base, approximately equal to \(2.71828\)

These properties make logarithmic functions particularly useful in calculus. For instance, in integration by parts, logarithms are often set as \(u\) because differentiating a logarithmic function simplifies the integrand and aids in solving complicated integration problems.

For the integral \\( \int x^n \ln(ax) \, dx \)\, the choice of \( u = \ln(ax) \) is strategic as it simplifies \(du = \frac{1}{x}dx\), making the whole process more manageable. Understanding these logarithmic properties, alongside mastering techniques like integration by parts, equips you to handle complex integrals effortlessly.