Problem 79 - Applied Calculus

Question

Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). \(\int \underbrace{x^{2} e^{x} d x}=x^{2} e^{x}-\int_{u} \underbrace{e^{x} 2 x d x}=x^{2} e^{x}-2 \int x e^{x} d x\) \(\left[\begin{array}{cc}u=x^{2} & d v=e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x}\end{array}\right]\) The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) : \(=x^{2} e^{x}-2\left(\int x e^{x} d x\right) \quad\left[\begin{array}{c}u=x \quad d v=e^{x} d x \\ d u=d x \quad v=e^{x}\end{array}\right]\) \(=x^{2} e^{x}-2\left(x e^{x}-\int e^{x} d x\right)\) \(=x^{2} e^{x}-2\left(x e^{x}-e^{x}\right)+C\) \(=x^{2} e^{x}-2 x e^{x}+2 e^{x}+C\) After reading the preceding explanation, find each integral by repeated integration by parts. \(\int x^{2}(\ln x)^{2} d x\)

Step-by-Step Solution

Answer

The integral is \( \frac{x^3}{3} (\ln x)^2 - \frac{2x^3}{9} (\ln x) + \frac{2x^3}{27} + C \).

Step 1: Choose First u and dv

In the integral \( \int x^2 (\ln x)^2 \, dx \), choose \( u = (\ln x)^2 \) and \( dv = x^2 \, dx \). Differentiate \( u \) to get \( du = 2(\ln x) \frac{1}{x} \, dx \) and integrate \( dv \) to get \( v = \frac{x^3}{3} \).

Step 2: Apply Integration by Parts Formula

Using the formula \( \int u \, dv = uv - \int v \, du \), we have:\[\int x^2 (\ln x)^2 \, dx = (\ln x)^2 \frac{x^3}{3} - \int \frac{x^3}{3} \, 2(\ln x) \frac{1}{x} \, dx\]Simplify to:\[\frac{x^3}{3} (\ln x)^2 - \frac{2}{3} \int x^2 (\ln x) \, dx\]

Step 3: Choose Second u and dv

For the integral \( \int x^2 (\ln x) \, dx \), choose new \( u = \ln x \) and \( dv = x^2 \, dx \). Differentiate \( u \) to get \( du = \frac{1}{x} \, dx \) and integrate \( dv \) to get \( v = \frac{x^3}{3} \).

Step 4: Apply Integration by Parts Again

Apply integration by parts once more:\[\int x^2 (\ln x) \, dx = \ln x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx\]Simplify to:\[\frac{x^3}{3} (\ln x) - \frac{1}{3} \int x^2 \, dx\]Since \( \int x^2 \, dx = \frac{x^3}{3} \), substitute:\[\frac{x^3}{3} (\ln x) - \frac{1}{9} x^3 + C_1\]

Step 5: Substitute Back and Simplify

Substitute back into the expression from Step 2:\[\frac{x^3}{3} (\ln x)^2 - \frac{2}{3} \left( \frac{x^3}{3} (\ln x) - \frac{1}{9} x^3 \right) + C\]Simplify the expression:\[\frac{x^3}{3} (\ln x)^2 - \frac{2x^3}{9} (\ln x) + \frac{2x^3}{27} + C\]

Step 6: Final Result

Combine and simplify all terms to achieve the final result:\[\frac{x^3}{3} (\ln x)^2 - \frac{2x^3}{9} (\ln x) + \frac{2x^3}{27} + C\]

Key Concepts

definite integrals

When dealing with definite integrals, you're essentially trying to find the area under the curve of a function between two specific points on the x-axis. In contrast to indefinite integrals, definite integrals result in a numeric value rather than a function plus a constant. Here’s how you tackle definite integrals:

• Define the region: Identify the function and the interval \([a, b]\) over which you're integrating. This interval is crucial as it bounds the area you aim to compute.
• Set up the integral: Write the integral of the function \(f(x)\) from \(a\) to \(b\). It looks like this: \[ \int_a^b f(x) \, dx \]
• Evaluate the antiderivative: Calculate the antiderivative, \(F(x)\), of the function \(f(x)\).
• Apply the Fundamental Theorem of Calculus: Evaluate \(F(b) - F(a)\) to find the absolute area under the curve between your chosen points on the x-axis.

Definite integrals are a powerful tool to calculate exact areas and are frequently used in physics, engineering, and probability.

indefinite integrals

Indefinite integrals help you find a general form of a function’s antiderivative. Unlike definite integrals, they do not have limits of integration and always include a constant of integration denoted as \(C\). Here is how you handle them:

• Recognize the function: Acknowledge the function you need to integrate.
• Integrate the function: Use known integration rules or techniques like substitution and integration by parts to integrate the function. This yields the antiderivative, \(F(x)\), plus an integration constant: \[ \int f(x) \, dx = F(x) + C \]
• Include the constant \(C\): Always remember to add \(C\) at the end of indefinite integrals since derivatives of constants are zero. \(C\) represents any constant offset, ensuring you encompass all possible antiderivatives.

Indefinite integrals are foundational in calculus, allowing you to calculate the original function from its rate of change or derivative. This process is often reversed when solving differential equations.

calculus techniques

There are several calculus techniques used to solve integrals, which help tackle various types of problems involving differentiation and integration. One pivotal technique is integration by parts.

**Integration by Parts**
Integration by parts is based on the product rule for differentiation. It is particularly useful for integrating products of functions and is expressed by the formula:
\[ \int u \, dv = uv - \int v \, du \]
Here’s a breakdown of the steps:

• Select \(u\) and \(dv\): Usually, \(u\) is chosen as a function that becomes simpler when differentiated, while \(dv\) is something easily integrable.
• Differentiate and integrate: Calculate \(du = u' \, dx\) and \(v = \int dv\).
• Substitute into the formula: Apply \[ \int u \, dv = uv - \int v \, du \]
• Simplify and repeat if needed: Some integrals may require multiple rounds of integration by parts, as seen in the exercise example with exponential and polynomial functions.

Other techniques include substitution, partial fractions, and trigonometric integrals, each tailored to specific forms of functions, diversifying your problem-solving arsenal in calculus.