Discovering the Velocity through Integration by Parts
A detailed exposition on solving and interpreting the calculus problem
Highlights
- Integration by parts is applied to evaluate the integral involved in modeling velocity.
- Both representations of the integrals, ∫e^x x dx and ∫x e^x dx, yield identical results.
- The method demonstrates that changing the order of factors in a product does not affect the outcome.
Introduction
In this detailed explanation, we explore the process of determining the velocity of an object whose acceleration is modeled by the expression \( \text{\text{a}} = e^x \, x \). The velocity function \( v \) is obtained through integration of the acceleration. Here, we use integration by parts to solve the integral \( v = \int e^x \, x \, dx \). Additionally, we examine whether the slight alternation in notation to \( v = \int x \, e^x \, dx \) affects the outcome of the problem. The explanation includes step-by-step procedures, derivations, and interpretations.
Part a) Determining the Velocity \( v = \int e^x x \, dx \)
Step-by-Step Process Using Integration by Parts
In this task, our goal is to integrate the function \( e^x x \) with respect to \( x \). The method of integration by parts is ideal here and is based on the formula:
Integration by Parts Formula
The formula is given by:
\( \displaystyle \int u \, dv = uv - \int v \, du \)
where \( u \) and \( dv \) are chosen from the given integrand.
Choosing \( u \) and \( dv \)
A careful choice of \( u \) and \( dv \) simplifies the process:
- Let \( u = x \) – its derivative will simplify the integral.
- Let \( dv = e^x \, dx \) – the integral of \( e^x \) is straightforward.
Determining \( du \) and \( v \)
With the choice above, differentiate and integrate accordingly:
- Differentiate: \( u = x \) hence \( du = dx \).
- Integrate: \( dv = e^x \, dx \) hence \( v = \int e^x \, dx = e^x \).
Applying the Integration by Parts Formula
Substitute the chosen parts into the formula:
\( \displaystyle \int e^x \, x \, dx = x \cdot e^x - \int e^x \, dx \)
Evaluating the Remaining Integral
The remaining integral \( \int e^x \, dx \) is simple:
\( \displaystyle \int e^x \, dx = e^x + C \)
where \( C \) represents the constant of integration.
Formulating the Final Expression
Incorporate the result from the evaluation:
\( \displaystyle \int e^x \, x \, dx = x \, e^x - \left( e^x + C \right) \)
Simplify by combining like terms:
\( \displaystyle \int e^x \, x \, dx = x \, e^x - e^x + C \)
or equivalently,
\( \displaystyle v = e^x (x - 1) + C \).
Part b) Assessing the Alternative Expression \( v = \int x \, e^x \, dx \)
Understanding the Commutative Property
The second part of the question asks if the problem changes when the integral is represented as \( \int x \, e^x \, dx \) rather than \( \int e^x \, x \, dx \). In both cases, the multiplicative terms \( e^x \) and \( x \) are simply being multiplied. Due to the commutative property of multiplication, which states that:
\( \displaystyle e^x \cdot x = x \cdot e^x \),
the order of the factors does not influence the outcome of the integral.
Applying the Integration Process Again
If we begin with \( v = \int x \, e^x \, dx \) and follow the same steps as before:
- Choose \( u = x \) and \( dv = e^x \, dx \).
- Then \( du = dx \) and \( v = e^x \).
Substituting in the integration by parts formula gives:
\( \displaystyle v = x \, e^x - \int e^x \, dx \).
Evaluating the remaining integral immediately leads to the same result:
\( \displaystyle \int e^x \, dx = e^x + C \)
So, once again, we have:
\( \displaystyle v = x \, e^x - e^x + C \)
or more neatly,
\( \displaystyle v = e^x (x - 1) + C \).
Conclusion on the Alternative Notation
The fundamental concept here is that the order of the product in the integrand does not affect the integration process or the final answer. Both forms, \( \int e^x \, x \, dx \) and \( \int x \, e^x \, dx \), yield the same integral:
\( \displaystyle v = e^x(x - 1) + C \).
It reaffirms that, in operations with multiplication, the arrangement of factors is interchangeable; hence, there is no difference in the nature of the problem or the resulting model for the velocity.
Detailed Analysis Summary
To ensure clarity, it’s beneficial to view the overall process in a structured manner. Below is a table summarizing each critical step of the problem-solving approach:
| Step | Action | Result |
|---|---|---|
| 1 | Identify the integral: \( \int e^x \, x \, dx \) | Target expression for velocity |
| 2 | Choose \( u = x \) and \( dv = e^x \, dx \) | Simplifies differentiation and integration |
| 3 | Compute \( du = dx \) and \( v = e^x \) | Obtained derivative and antiderivative |
| 4 | Apply integration by parts: \( uv - \int v \, du \) | Expression becomes \( x \, e^x - \int e^x \, dx \) |
| 5 | Simplify \( \int e^x \, dx \) | Results in: \( e^x + C \) |
| 6 | Subtract and combine terms | Final integrated result: \( e^x (x-1) + C \) |
| 7 | Examine alternative form \( \int x \, e^x \, dx \) | Same procedure and same outcome |
The table above encapsulates the logical flow from the initial integral to the final expression. It demonstrates that the process of selecting the parts, applying the formula, and evaluating the simple exponential integral is consistent regardless of minor changes in the notation.
In-depth Explanation of Key Concepts
Integration by Parts
Integration by parts is a powerful technique in calculus used specifically for integrating products of functions. The method essentially transforms a complicated integral into a simpler one. A crucial insight here is recognizing which function to differentiate (making it simpler) and which to integrate (keeping it manageable). In this scenario, choosing \( u = x \) is practical because its derivative is 1, while \( dv = e^x \, dx \) is equally convenient since the antiderivative of \( e^x \) is itself.
The formula itself, \( \displaystyle \int u\,dv = uv - \int v\,du \), is applied systematically to help reduce the integration process into simpler evaluations, making it particularly useful in cases with products where one component simplifies significantly upon differentiation.
Order of Multiplication and Its Non-effect on Integration
The question also emphasizes understanding properties of multiplication in integrals. Specifically, whether writing the integrand as \( e^x \, x \) or \( x \, e^x \) changes the outcome. The principle at play here is the commutative property of multiplication, which states that the order of factors does not impact the product. Consequently, both expressions represent the same mathematical quantity and yield identical integrals.
This verification is crucial because it not only simplifies the problem but also confirms that the method and results are robust under slight changes in the representation, reinforcing the consistency of calculus rules.
Additional Explanatory Insights
Handling the Constant of Integration
It is important to note that during any indefinite integration, the constant of integration \( C \) is added to the final result. This constant represents an infinite family of antiderivatives that differ only by a constant. In our specific problem, the inclusion of \( C \) in \( v = e^x (x-1) + C \) ensures that the solution accounts for all potential antiderivatives, making the expression general.
Practical Applications
The methodology described here is not limited to the current example. Integration by parts is widely used in physics and engineering, especially when modeling dynamic systems where acceleration, velocity, and displacement are interconnected through differential equations. For example, similar integration techniques appear in problems involving damped harmonic motion, electrical circuits with exponential behaviors, and even in probability theory when handling moment-generating functions.
Mastery over such techniques is crucial for solving real-world problems that require the integration of complex functions.
References
- Integration by Parts Example - Petervis
- Integration by Parts - Brilliant
- Integral of \(x e^x\) - Cuemath
- Integration by Parts - Maths is Fun
- Integration Calculator - Symbolab
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Last updated March 24, 2025