Problem 78 - 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. \(\int(\ln x)^{2} d x\)

Step-by-Step Solution

Answer

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

Step 1: Choose Initial u and dv

We start with the integral \( \int (\ln x)^{2} \, dx \). Let's choose \( u = (\ln x)^{2} \) and \( dv = dx \). This implies \( du = 2 \ln x \cdot \frac{1}{x} \, dx = \frac{2 \ln x}{x} \, dx \) and \( v = x \).

Step 2: Apply Integration by Parts Formula

Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:\[\begin{align*}\int (\ln x)^{2} \, dx &= x \cdot (\ln x)^{2} - \int x \cdot \frac{2 \ln x}{x} \, dx \&= x (\ln x)^{2} - 2 \int \ln x \, dx\end{align*}\]

Step 3: Solve the Remaining Integral

Now we need to solve the integral \( \int \ln x \, dx \) using integration by parts again. Choose \( u = \ln x \) and \( dv = dx \), which gives \( du = \frac{1}{x} \, dx \) and \( v = x \).Using integration by parts, \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C' \).

Step 4: Substitute Back into Original Equation

Substitute \( \int \ln x \, dx \) back into the original equation:\[\begin{align*}x (\ln x)^{2} - 2 \int \ln x \, dx &= x (\ln x)^{2} - 2 (x \ln x - x + C') \&= x (\ln x)^{2} - 2x \ln x + 2x - 2C'\end{align*}\]

Step 5: Final Simplification

Combine and simplify the parts:\[x (\ln x)^{2} - 2x \ln x + 2x + C\]Thus, the integration yields the result.

Key Concepts

repeated integration

Repeated integration by parts is a powerful technique when solving integrals involving functions that can be decomposed into products. This method is particularly helpful when an initial integration by parts leads to another integral that can also be tackled using the same method.

To effectively use repeated integration by parts, remember these key steps:

• Identify suitable parts of the integrand to choose as \( u \) and \( dv \).
• Calculate \( du \) and integrate \( dv \) to get \( v \).
• Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
• Check if the remaining integral can be solved directly or by further integration by parts.

In the original exercise with \( \int x^2 e^x \, dx \), after the first integration by parts, we are left with another integral \( \int xe^x \, dx \). Here, a second integration by parts simplifies the problem further until the entire expression is integral-free.

definite integral

While the exercise focuses on indefinite integrals, understanding definite integrals is crucial for mathematics and its applications. Definite integrals are similar to indefinite ones, but have specific limits of integration, providing a numerical value that represents the area under a curve between two points.

When using integration by parts for a definite integral, extra attention is needed for the boundaries. The integration by parts formula for definite integrals is:

• \( \int_a^b u \, dv = \left[uv\right]_a^b - \int_a^b v \, du \)

For example, if we had specific limits \(a\) and \(b\) for \( \int_a^b x^2 e^x \, dx \), we must evaluate each term from the integration by parts at these limits. This approach illuminates why integrals are not just algebraic but also geometric tools.

integration techniques

Mastering integration means becoming familiar with various techniques that simplify or transform difficult integrals into manageable forms. Apart from integration by parts, some other essential techniques are:

• Substitution: Useful when an integral contains a composite function. Replace a section of the function with a single variable to simplify.
• Partial Fractions: Effective for rational functions where the numerator degree is less than the denominator's.
• Trigonometric Integrals: Apply identities to change products or powers of trigonometric functions into simpler forms.
• Numerical Methods: When analytical solutions are difficult, numerical approximation methods like Simpson's rule or the trapezoidal rule can be helpful.

Each technique has its strengths and optimal use cases, enabling you to tackle a wide range of integration problems more efficiently.

calculus problem solving

Solving calculus problems, especially with integrals, often requires a strategic approach, understanding the problem, and selecting the most suitable method. To effectively solve calculus problems, consider these steps:

• Understand the Problem: Carefully read the problem and identify what needs to be calculated or proved.
• Select a Method: Choose an integration technique that aligns with the problem's form and structure. Look for clues in the integrand that suggest particular methods.
• Break Down Complex Problems: For complex integrals like the one in the exercise, breaking them down into smaller integrals using repeated integration by parts helps simplify the task.
• Verify Solutions: Check your work by differentiating your result to see if it returns to the original integrand.

This approach ensures a clear path from integral to solution, promoting a deeper understanding of calculus and its applications.