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Magazine Chapter 28: Problem 5
Integrate each of the given functions. $$\int 4 x e^{2 x} d x$$
Short Answer
Expert verified
The integral is \( 2x e^{2x} - e^{2x} + C \).
Step by step solution
01
Identify the Integration Technique
The integral \( \int 4xe^{2x} dx \) requires the use of integration by parts. Integration by parts is based on the formula: \[\int u \, dv = uv - \int v \, du\]We need to choose \( u \) and \( dv \) from our integral.
02
Choose Parts for Integration by Parts
Choose \( u = 4x \) and \( dv = e^{2x} \, dx \). Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \):\[u = 4x \quad \Rightarrow \quad du = 4 \, dx\]\[dv = e^{2x} \, dx \quad \Rightarrow \quad v = \frac{1}{2}e^{2x}\]
03
Apply the Integration by Parts Formula
Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[\int 4x e^{2x} \, dx = uv - \int v \, du = 4x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \, 4 \, dx\]Simplify the expression to continue: \[= 2x e^{2x} - 2 \int e^{2x} \, dx\]
04
Integrate the Remaining Integral
Now, integrate \( \int e^{2x} \, dx \):\[\int e^{2x} \, dx = \frac{1}{2}e^{2x}\]
05
Complete the Integration
Substitute the integral result back into the expression:\[2x e^{2x} - 2 \cdot \frac{1}{2}e^{2x}\]This simplifies to:\[2x e^{2x} - e^{2x}\]
06
Include the Constant of Integration
Add the constant of integration \( C \) to complete the indefinite integral:\[\int 4x e^{2x} \, dx = 2x e^{2x} - e^{2x} + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Techniques of Integration
Integration is a powerful tool in calculus used to calculate areas, volumes, and other quantities. For more complex functions like the one in this problem, techniques such as *integration by parts* become necessary. Integration by parts is a specific method where the product of two functions is integrated. The formula used is:
In this formula, \( u \) and \( dv \) are chosen from the integrand. The derivative \( du \) and integral \( v \) are then calculated.
This technique is particularly useful when you face products of polynomial and exponential functions, like \( 4x e^{2x} \). By smartly choosing \( u \) and \( dv \), we simplify the problem into easier parts to integrate.
- \( \int u \, dv = uv - \int v \, du \)
In this formula, \( u \) and \( dv \) are chosen from the integrand. The derivative \( du \) and integral \( v \) are then calculated.
This technique is particularly useful when you face products of polynomial and exponential functions, like \( 4x e^{2x} \). By smartly choosing \( u \) and \( dv \), we simplify the problem into easier parts to integrate.
Indefinite Integrals
An indefinite integral is simply the reverse process of differentiation. It finds a function whose derivative is the given function. In this context, indefinite integrals do not have specific limits, unlike definite integrals.
When performing indefinite integration, it鈥檚 crucial to include a constant of integration, denoted as \( C \). This constant is essential because integration can yield multiple functions differing by a constant factor:
In simpler terms, it underscores that each antiderivative is just one part of a family of functions that differ by a constant.
When performing indefinite integration, it鈥檚 crucial to include a constant of integration, denoted as \( C \). This constant is essential because integration can yield multiple functions differing by a constant factor:
- \( \int f(x) \, dx = F(x) + C \)
In simpler terms, it underscores that each antiderivative is just one part of a family of functions that differ by a constant.
Calculus Problem-Solving
Solving calculus problems often requires a mix of knowledge, strategy, and practice. When approaching a problem, it's helpful to:
Embracing these strategies can not only aid in solving problems faster but also deepen understanding and foster confidence in calculus.
- Identify the type of function and technique needed for integration.
- Break down the problem into manageable parts using appropriate methods like substitution or integration by parts.
- Simplify expressions as much as possible to make them easier to integrate.
Embracing these strategies can not only aid in solving problems faster but also deepen understanding and foster confidence in calculus.
