Problem 50 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(47-50,\) use integration by parts to establish the reduction formula. $$\int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x$$

Step-by-Step Solution

Answer

The reduction formula for the given integral \( \int (\ln x)^n dx \) is \( x(\ln x)^n - n \int (\ln x)^{n-1} dx \).

Step 1: Identify Functions u and dv

First thing in integration by parts is to choose the functions u and dv. In this case, choose \( u=(\ln x)^n \) and \( dv=dx \).

Step 2: Compute the Derivative and Integral

Calculate the derivative of u (du) and the integral of dv (v). The derivative of \( u=(\ln x)^n \) is \( du = n (\ln x)^{n-1} dx / x \). And the integral of \( dx \) is simply \( v = x \).

Step 3: Apply the Integration by Parts Formula

Substitute u, v, du, and dv in the integration by parts formula \(\int udv = uv - \int vdu\). This leads to: \(\int (\ln x)^n dx = x(\ln x)^n - \int x * n (\ln x)^{n-1} / x dx\). Simplify this to get the desired reduction formula: \(\int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx\).

Key Concepts

Reduction Formula

The concept of a reduction formula is a powerful tool in calculus, especially when dealing with integration. It simplifies the process of solving complex integrals by reducing the problem to the integration of a function with a smaller power or a simpler form. The exercise shows how integration by parts can be used to derive such a formula for the integral of \(\ln x\)^n.

Integration by parts is based on the product rule for differentiation and connects the integral of a product of two functions to the integral of their derivatives. The exercise takes you through identifying the functions \(u\) and \(dv\), calculating their derivatives and antiderivatives, and then applying the integration by parts formula to arrive at the reduction formula. This process showcases a pattern that, once recognized, can be applied recursively to integrate functions raised to any power \(n\), by successively reducing the exponent until a base case is reached.

Logarithmic Integration

When facing an integral that includes a logarithmic function such as \(\ln x\), a technique known as logarithmic integration is often useful. This approach typically involves using integration by parts, where the logarithmic function and its derivative play key roles. In the given exercise, the choice of \(u = (\ln x)^n\) allowed the application of this technique, where the integral of the logarithmic function is incrementally reduced.

Why Logarithmic Integration?

Logarithmic integration is particularly advantageous when dealing with powers of natural logarithms, as the derivative of \(\ln x\) leads to \(1/x\), a simpler expression that is easier to integrate. It effectively decomposes a more complex expression into more manageable parts, leveraging the inherent properties of logarithms to simplify the integration process.

Definite Integration

The process of definite integration extends beyond finding antiderivatives; it is about evaluating the area under the curve of a function between two specified points. The reduction formula derived in this exercise can be used to calculate these areas when dealing with definite integrals involving logarithmic functions. Here's how to apply it:

When provided with limits of integration, you would first use the reduction formula to express the integral in a simpler form, then evaluate the simplified integral at the upper and lower bounds and subtract. This is an essential process in real-world applications, as finding the definite integral represents tangible quantities like distances, areas, and volumes. Enhancing the step-by-step solution with this context can help students appreciate how definite integration serves as a bridge between abstract mathematical concepts and practical problems.