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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.2.10

Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ (x² - 2x + 1) e^(2x) dx

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1
Identify the integral to solve: \(\int (x^{2} - 2x + 1) e^{2x} \, dx\).
Recall the integration by parts formula: \(\int u \, dv = uv - \int v \, du\).
Choose \(u\) and \(dv\) wisely. Let \(u = (x^{2} - 2x + 1)\) because its derivative simplifies the polynomial, and let \(dv = e^{2x} \, dx\) because the integral of an exponential function is straightforward.
Compute \(du\) and \(v\): differentiate \(u\) to get \(du = (2x - 2) \, dx\), and integrate \(dv\) to get \(v = \frac{1}{2} e^{2x}\).
Apply the integration by parts formula: write \(\int (x^{2} - 2x + 1) e^{2x} \, dx = (x^{2} - 2x + 1) \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (2x - 2) \, dx\), then simplify and prepare to integrate the remaining integral, which may require applying integration by parts again.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the problem, especially when one function becomes simpler upon differentiation.
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Polynomial and Exponential Functions

Understanding the behavior of polynomial functions like x² - 2x + 1 and exponential functions like e^(2x) is crucial. Polynomials simplify upon differentiation, while exponentials remain similar upon integration and differentiation, making them suitable candidates for integration by parts.
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Repeated Application of Integration by Parts

When integrating products involving polynomials and exponentials, multiple iterations of integration by parts may be necessary. Each step reduces the polynomial degree, eventually leading to an integral that can be directly evaluated, ensuring the problem is fully solved.
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