Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 10
Evaluate the integrals in Exercises \(1-24\) using integration by parts. $$ \int\left(x^{2}-2 x+1\right) e^{2 x} d x $$
Short Answer
Expert verified
The integral evaluates to \( \frac{e^{2x}}{2}x^2 - e^{2x}x + \frac{3e^{2x}}{4} + C \).
Step by step solution
01
Choose Parts
Identify the two parts of the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Let \( u = x^2 - 2x + 1 \) (polynomial) and \( dv = e^{2x} \, dx \) (exponential).
02
Differentiate and Integrate
Differentiate \( u \) to find \( du \):\[ du = (2x - 2) \, dx \] Integrate \( dv \) to find \( v \):\[ v = \frac{1}{2}e^{2x} \]
03
Substitute Into the Formula
Apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substitute \( u, v, \) and \( du \) into the equation: \[ \int (x^2 - 2x + 1)e^{2x} \ dx = (x^2 - 2x + 1) \left( \frac{1}{2}e^{2x} \right) - \int \left( \frac{1}{2}e^{2x} \right)(2x - 2) \ dx \] Simplify: \[ = \frac{(x^2 - 2x + 1)e^{2x}}{2} - \int e^{2x}(x - 1) \ dx \]
04
Apply Integration by Parts Again
Now tackle \( \int e^{2x}(x - 1) \, dx \) using integration by parts again. Let \( u = x - 1 \) and \( dv = e^{2x} \, dx \). Differentiating and integrating, we have \( du = dx \) and \( v = \frac{1}{2}e^{2x} \).
05
Second Integration by Parts
Apply the integration by parts formula to \( \int e^{2x}(x - 1) \, dx \): \[ \int (x - 1)e^{2x} \ dx = \frac{x - 1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \ dx \] The remaining integral is \( \frac{1}{2} \int e^{2x} \, dx = \frac{1}{4} e^{2x} \). Combine terms: \[ \int (x - 1)e^{2x} \, dx = \frac{(x - 1)e^{2x}}{2} - \frac{e^{2x}}{4} \]
06
Combine Results
Replace \( \int e^{2x}(x - 1) \, dx \) in the main expression: \[ \int (x^2 - 2x + 1)e^{2x} \, dx = \frac{e^{2x}(x^2 - 2x + 1)}{2} - \left( \frac{(x - 1)e^{2x}}{2} - \frac{e^{2x}}{4} \right) \] Simplify and combine the terms: \[ = \frac{e^{2x}}{2}x^2 - e^{2x}x + \frac{3e^{2x}}{4} + C \]
07
Final Answer
The integral is evaluated as: \[ \int (x^2 - 2x + 1)e^{2x} \, dx = \frac{e^{2x}}{2}x^2 - e^{2x}x + \frac{3e^{2x}}{4} + C \] where \( C \) is the constant of integration.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a powerful concept in calculus that let us calculate the exact area under a curve. They are written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
In the context of integration by parts, definite integrals play an essential role when determining accumulated quantities over a certain interval.
Let's break down why they are important:
In the context of integration by parts, definite integrals play an essential role when determining accumulated quantities over a certain interval.
Let's break down why they are important:
- **Area Calculation:** Definite integrals give the exact area under the curve of a function from \( a \) to \( b \).
- **Applications:** Useful in fields like physics (for work or force over a distance), economics (consumer and producer surplus), and more.
- **Integration by Parts with Limits:** In problems with limits, after applying integration by parts, remember to evaluate the result using the original limits \( a \) and \( b \).
Exponential Functions
Exponential functions are integral to many calculus problems, especially in physics and natural sciences. They are written as \( e^{kx} \), where \( e \) is a constant approximately equal to 2.71828, and \( k \) is a constant that determines the function's growth rate.
In the integration by parts method, dealing with exponential functions can simplify the process because they are their own derivatives. Here's why:
In the integration by parts method, dealing with exponential functions can simplify the process because they are their own derivatives. Here's why:
- **Derivative and Integral:** The derivative of \( e^{kx} \) is \( ke^{kx} \), and its integral is \( \frac{1}{k}e^{kx} \). This makes exponential functions very special and easy to integrate.
- **Growth Factor:** The function \( e^{kx} \) grows or decays rapidly, depending on the sign of \( k \). This makes exponential functions significant in modeling exponential growth or decay processes.
- **Base Formula Usage:** In the problem, we use the exponential function to find \( v \) by integrating \( dv \), which results in \( v = \frac{1}{2}e^{2x} \), leveraging the property of exponential functions for elegant solutions.
Polynomial Integration
Polynomial integration involves finding a function whose derivative gives you the original polynomial. This is a common task in calculus and is crucial for solving complex integrals.
When it comes to integration by parts, polynomials are frequently employed as one of the chosen parts. Here's the breakdown of how it works:
When it comes to integration by parts, polynomials are frequently employed as one of the chosen parts. Here's the breakdown of how it works:
- **Structure of Polynomials:** Polynomials have the form \( ax^n + bx^{n-1} + ... + c \), where \( n \) is a positive integer. In integration by parts, typically, the polynomial is chosen as \( u \) because differentiating makes it simpler.
- **Reduction Method:** By differentiating \( u \), the degree of the polynomial reduces, which gradually simplifies the integrand during each integration by parts iteration.
- **Example from Exercise:** In the given exercise, the polynomial \( x^2 - 2x + 1 \) was selected as \( u \). When differentiated, it resulted in \( du = (2x - 2) \, dx \), streamlining the overall integration process.
