Integration Of X 2 Ex

Integrating x²eˣ: A Comprehensive Guide

The integral of x²eˣ is a classic example of integration by parts, a technique frequently encountered in calculus. This seemingly simple integral hides a surprisingly elegant solution and provides a fantastic opportunity to explore the power and versatility of integration techniques. This comprehensive guide will not only show you how to solve this integral but also delve into the underlying principles and provide a deeper understanding of the method. We'll explore the steps involved, examine the underlying mathematical concepts, and even answer some frequently asked questions. Mastering this integral will significantly enhance your calculus skills and pave the way for tackling more complex integration problems.

Understanding Integration by Parts

Before we dive into the specific integral of x²eˣ, let's refresh our understanding of integration by parts. This technique is a direct consequence of the product rule for differentiation. Recall that the product rule states:

d(uv)/dx = u(dv/dx) + v(du/dx)

Integrating both sides with respect to x, we get:

∫d(uv)/dx dx = ∫u(dv/dx) dx + ∫v(du/dx) dx

This simplifies to:

uv = ∫u(dv/dx) dx + ∫v(du/dx) dx

Rearranging this equation, we arrive at the integration by parts formula:

∫u(dv/dx) dx = uv - ∫v(du/dx) dx

This formula allows us to solve integrals of products of functions. The key is choosing appropriate 'u' and 'dv/dx' such that the integral on the right-hand side is simpler to evaluate than the original integral.

Step-by-Step Integration of x²eˣ

Now, let's tackle the integral of x²eˣ. We'll apply integration by parts twice.

Step 1: The First Application of Integration by Parts

We need to choose our 'u' and 'dv/dx'. A good rule of thumb is to choose 'u' as the function that simplifies when differentiated and 'dv/dx' as the function that is easily integrated. In this case:

• u = x² (This simplifies when differentiated)
• dv/dx = eˣ (This is easily integrated)

Now we find du/dx and v:

• du/dx = 2x
• v = eˣ

Substituting these into the integration by parts formula:

∫x²eˣ dx = x²eˣ - ∫eˣ(2x) dx

Notice that we've reduced the power of x from 2 to 1. We still have an integral to solve, but it's simpler than the original.

Step 2: The Second Application of Integration by Parts

We now need to solve the integral ∫2xeˣ dx. We'll apply integration by parts again:

• u = 2x
• dv/dx = eˣ

Therefore:

• du/dx = 2
• v = eˣ

Applying the integration by parts formula again:

∫2xeˣ dx = 2xeˣ - ∫eˣ(2) dx

This integral, ∫2eˣ dx, is straightforward:

∫2eˣ dx = 2eˣ + C (where C is the constant of integration)

Step 3: Combining the Results

Now we substitute this result back into the equation from Step 1:

∫x²eˣ dx = x²eˣ - (2xeˣ - 2eˣ + C)

Simplifying, we get the final result:

∫x²eˣ dx = x²eˣ - 2xeˣ + 2eˣ + C

A Deeper Look: The Pattern and Generalization

Notice the pattern in the solution: The powers of x decrease by one with each application of integration by parts, and the coefficients follow a factorial pattern. This observation allows us to generalize the integral of xⁿeˣ:

∫xⁿeˣ dx = xⁿeˣ - nxⁿ⁻¹eˣ + n(n-1)xⁿ⁻²eˣ - ... + (-1)ⁿn!eˣ + C

This formula can be proven using mathematical induction, but it's clear from our example with x²eˣ.

Why This Method Works: A Mathematical Perspective

The success of integration by parts in this case hinges on the fact that the derivative of xⁿ (a polynomial) eventually becomes zero. Each application of integration by parts reduces the power of x, ultimately leading to a simple integral that can be easily solved. The exponential function, eˣ, remains unchanged under differentiation and integration, making it a suitable partner for integration by parts with polynomial functions.

Practical Applications

The integral of x²eˣ, and its generalizations, appear frequently in various fields, including:

• Physics: Solving differential equations related to exponential decay or growth, particularly in problems involving time-dependent forces or quantities.
• Probability and Statistics: Evaluating expectations and moments of probability distributions.
• Engineering: Solving problems in electrical circuits, signal processing, and control systems involving exponential functions.

Frequently Asked Questions (FAQ)

Q1: Can I choose 'u' and 'dv/dx' differently?

A1: Yes, but the choice of 'u' and 'dv/dx' significantly impacts the complexity of the subsequent integration. While other choices are possible, the approach outlined above is generally the most efficient for this particular integral.

Q2: What if the exponent wasn't 'x' but a more complex expression?

A2: If the exponent were a more complex expression, say ax + b, you might need to use substitution before applying integration by parts. The core principle of integration by parts remains the same, but the steps might become more involved.

Q3: What if I have x²e^(ax²) instead of x²eˣ?

A3: This integral cannot be solved using elementary functions. It requires more advanced techniques like using series expansions or numerical methods.

Q4: Are there other methods to solve this integral?

A4: While integration by parts is the most straightforward and efficient approach, other methods, such as using Laplace transforms or generating functions, could be employed but are generally more complex for this specific problem.

Q5: Why is the constant of integration 'C' important?

A5: The constant of integration 'C' represents the family of functions whose derivative is x²eˣ. Omitting it results in an incomplete and inaccurate solution.

Conclusion

Integrating x²eˣ provides a valuable learning experience in applying integration by parts effectively. By systematically reducing the power of the polynomial term and utilizing the properties of the exponential function, we arrive at a concise and elegant solution. Understanding this process not only solves a specific integral but also equips you with a powerful technique for tackling a wide range of integration problems. Remember the steps, understand the underlying mathematical principles, and practice regularly to master this essential calculus skill. The ability to confidently integrate functions like x²eˣ opens doors to more advanced concepts and applications in various scientific and engineering disciplines.