Problem 67
Question
Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int x^{3} e^{2 x} d x$$
Step-by-Step Solution
Verified Answer
Question: Evaluate the integral: \(\int x^3 e^{2x} dx\)
Answer: \(\int x^3 e^{2x} dx = \frac{1}{2} x^{3} e^{2x} - \frac{3}{4}x^{2} e^{2x} + \frac{3}{4}x e^{2x} - \frac{3}{8} e^{2x} + C\)
1Step 1: Choose u and dv
First, let's apply the integration by parts formula, which states that:
$$\int u dv = uv - \int v du$$
Choose \(u = x^{n}\) and \(dv = e^{\alpha x}\), where \(n = 3\) and \(\alpha = 2\). Next, differentiate \(u\) to obtain \(du\) and integrate \(dv\) to obtain \(v\).
$$du = \frac{d}{dx} x^{n} = nx^{n-1}dx$$
$$v = \int e^{\alpha x} dx = \frac{1}{\alpha} e^{\alpha x}$$
2Step 2: Apply integration by parts
Plug the chosen values of \(u\), \(v\), \(du\), and \(dv\) into the original formula:
$$\int x^{3} e^{2 x} dx = x^{3} \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} (3x^{2}) dx = \frac{1}{2}x^{3} e^{2x} - \frac{3}{2}\int x^{2}e^{2x}dx$$
3Step 3: Apply integration by parts again
Repeat the process of choosing \(u\), \(dv\), differentiating and integrating for the remaining integral, using now \(u = x^{2}\), \(dv = e^{2x}\):
$$du = 2x dx$$
$$v = \frac{1}{2} e^{2x}$$
And apply the integration by parts formula:
$$\int x^{2} e^{2 x} dx = \frac{1}{2}x^{2} e^{2x}- \int x e^{2x}dx$$
Substitute back into the first integral result to get:
$$\int x^{3} e^{2 x} dx = \frac{1}{2} x^{3} e^{2x} -\frac{3}{2} \left( \frac{1}{2}x^{2} e^{2x}- \int x e^{2x}dx \right)$$
4Step 4: Apply integration by parts for the last time
Again, apply the integration by parts formula for the remaining integral, using \(u = x\), \(dv = e^{2x}\):
$$du = dx$$
$$v = \frac{1}{2} e^{2x}$$
And apply the integration by parts formula:
$$\int x e^{2 x} dx = \frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x} dx = \frac{1}{2}x e^{2x} - \frac{1}{4} e^{2x}$$
Finally, substitute the last result back into the second integral result:
$$\int x^{3} e^{2 x} dx = \frac{1}{2} x^{3} e^{2x} -\frac{3}{2} \left( \frac{1}{2}x^{2} e^{2x}- \left(\frac{1}{2}x e^{2x} - \frac{1}{4} e^{2x}\right) \right)$$
5Step 5: Simplify the result
Eliminate parentheses and collect like terms to obtain the final answer:
$$\int x^{3} e^{2 x} dx = \frac{1}{2} x^{3} e^{2x} - \frac{3}{4}x^{2} e^{2x} + \frac{3}{4}x e^{2x} - \frac{3}{8} e^{2x} + C$$
Where \(C\) is the constant of integration.
Key Concepts
Reduction FormulasDefinite IntegralsExponential FunctionsCalculus Problem-Solving
Reduction Formulas
Reduction formulas are a handy tool in calculus, especially when dealing with complex integrals. They provide a way to express an integral involving powers of variables in terms of integrals of lower powers, effectively reducing the complexity of a problem.
When performing integration by parts repeatedly, like in our exercise above, the reduction formulas guide the simplification process. They help in finding a pattern or a sequence that brings the integral to a simpler form.
This is particularly useful for integrals involving polynomial expressions combined with exponential functions, as they often require multiple applications of integration by parts. By using reduction formulas, you can streamline your work and avoid repetitive calculations.
When performing integration by parts repeatedly, like in our exercise above, the reduction formulas guide the simplification process. They help in finding a pattern or a sequence that brings the integral to a simpler form.
This is particularly useful for integrals involving polynomial expressions combined with exponential functions, as they often require multiple applications of integration by parts. By using reduction formulas, you can streamline your work and avoid repetitive calculations.
Definite Integrals
Definite integrals compute the area under a curve bounded by a specific interval on the x-axis. They provide a fixed numerical value as opposed to indefinite integrals, which yield a general expression plus a constant.
In this exercise, while we focused on finding an indefinite integral, the concept of definite integrals is essential for understanding how these calculations produce tangibly useful results.
Definite Integrals involve a few key properties:
In this exercise, while we focused on finding an indefinite integral, the concept of definite integrals is essential for understanding how these calculations produce tangibly useful results.
Definite Integrals involve a few key properties:
- The Fundamental Theorem of Calculus, which connects differentiation and integration.
- Linearity property allows us to split integrals over addition/subtraction.
- The evaluation over a boundary since the result represents the net area over an interval.
Exponential Functions
Exponential functions are fundamental in mathematics, characteristically expressed as \(e^x\), with important properties of growth and decay. They have unique mathematical attributes, like the fact that their derivative and integral processes are self-similar.
In calculus, exponential functions often appear in integrals, alongside polynomials, which makes them a prime candidate for the application of integration by parts.
The function \(e^{kx}\) is versatile and finds applications in fields involving compound interest, population growth, radioactive decay, etc. Its behavior remains consistent, preserving its structure, which simplifies calculations compared to other more varied functions.
Understanding how to manipulate these functions in integration is critical.
In calculus, exponential functions often appear in integrals, alongside polynomials, which makes them a prime candidate for the application of integration by parts.
The function \(e^{kx}\) is versatile and finds applications in fields involving compound interest, population growth, radioactive decay, etc. Its behavior remains consistent, preserving its structure, which simplifies calculations compared to other more varied functions.
Understanding how to manipulate these functions in integration is critical.
Calculus Problem-Solving
Tackling calculus problems requires a strong understanding of both the concepts and the methods used to solve challenging integrals. The use of integration by parts involves strategy: discriminately choosing which part of the integral to differentiate or integrate.
This exercise demonstrates how breaking down a large integral into manageable sections makes the process more straightforward. Key strategies include:
This exercise demonstrates how breaking down a large integral into manageable sections makes the process more straightforward. Key strategies include:
- Carefully selecting "u" and "dv" to ensure easier subsequent integrations or differentiations.
- Repeating the process to simplify progressively.
- Constantly checking work to notice patterns or areas for simplification.
Other exercises in this chapter
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