Problem 37
Question
Use integration by parts to verify the validity of the reduction formula $$ \int x^{n} e^{a x} d x=\frac{1}{a} x^{n} e^{a x}-\frac{n}{a} \int x^{n-1} e^{a x} d x $$ where \(a\) is a constant not equal to 0 . (b) Apply the reduction formula in (a) to compute $$ \int x^{2} e^{-3 x} d x $$
Step-by-Step Solution
Verified Answer
\( \int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} - \frac{2}{9} x e^{-3x} - \frac{2}{27} e^{-3x} \)
1Step 1: Set Up Integration By Parts Formula
The integration by parts formula is \[ \int u \ dv = uv - \int v \ du \]For this problem, choose \( u = x^n \) and \( dv = e^{ax} \, dx \). Then, \( du = n x^{n-1} \, dx \) and \( v = \frac{1}{a} e^{ax} \).
2Step 2: Apply Integration By Parts
Using integration by parts, substitute the expressions for \( u, v, du, \) and \( dv \) into the formula:\[ \int x^n e^{ax} \, dx = \frac{1}{a} x^n e^{ax} - \int \left( \frac{1}{a} e^{ax} \right) n x^{n-1} \, dx \]This simplifies to:\[ \int x^n e^{ax} \, dx = \frac{1}{a} x^n e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \]This confirms the reduction formula presented in the given problem.
3Step 3: Apply the Reduction Formula
We want to compute \( \int x^2 e^{-3x} \, dx \) using the reduction formula:\[ \int x^n e^{ax} \, dx = \frac{1}{a} x^n e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \]Substitute \( n = 2 \) and \( a = -3 \):\[ \int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} - \frac{2}{-3} \int x e^{-3x} \, dx \]Simplifies to:\[ \int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} + \frac{2}{3} \int x e^{-3x} \, dx \]
4Step 4: Apply Reduction Formula Again
Apply the reduction formula for \( \int x e^{-3x} \, dx \) by setting \( n = 1 \) and \( a = -3 \):\[ \int x e^{-3x} \, dx = -\frac{1}{3} x e^{-3x} - \frac{1}{-3} \int e^{-3x} \, dx \]This further simplifies to:\[ \int x e^{-3x} \, dx = -\frac{1}{3} x e^{-3x} + \frac{1}{3} \cdot \frac{-1}{3} e^{-3x} = -\frac{1}{3} x e^{-3x} - \frac{1}{9} e^{-3x} \]
5Step 5: Simplify and Combine Terms
Substitute the result from Step 4 back into the expression from Step 3:\[ \int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} + \frac{2}{3} \left(-\frac{1}{3} x e^{-3x} - \frac{1}{9} e^{-3x} \right) \]Simplify:\[ \int x^2 e^{-3x} \, dx = -\frac{1}{3} x^2 e^{-3x} - \frac{2}{9} x e^{-3x} - \frac{2}{27} e^{-3x} \]
Key Concepts
Reduction FormulaIntegration TechniquesIndefinite Integrals
Reduction Formula
The reduction formula is a vital tool in integration, especially when dealing with complex expressions. By simplifying the process of evaluating integrals, you can systematically reduce a complicated integral into simpler forms. The power of the reduction formula lies in its ability to repeatedly apply integration until the expression becomes manageable.
Consider the formula provided: \[ \int x^{n} e^{ax} \, dx = \frac{1}{a} x^{n} e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \]What’s happening here is an elegant breakdown of the integral into a core component and a reduced version of itself. This allows the integral involving the higher power \(x^n\) to transform into one with a lower power \(x^{n-1}\). Progressively repeating this reduces the problem to a basic integral which is solvable using standard techniques.
This recursive application not only helps in simplifying complex problems but is also instrumental in deriving new relationships and integrations that would be otherwise challenging to compute directly.
Consider the formula provided: \[ \int x^{n} e^{ax} \, dx = \frac{1}{a} x^{n} e^{ax} - \frac{n}{a} \int x^{n-1} e^{ax} \, dx \]What’s happening here is an elegant breakdown of the integral into a core component and a reduced version of itself. This allows the integral involving the higher power \(x^n\) to transform into one with a lower power \(x^{n-1}\). Progressively repeating this reduces the problem to a basic integral which is solvable using standard techniques.
This recursive application not only helps in simplifying complex problems but is also instrumental in deriving new relationships and integrations that would be otherwise challenging to compute directly.
Integration Techniques
Integration by parts is one of several essential techniques used in calculus to solve integral problems. This particular method is especially useful when the integrand is a product of two functions, like \(x^n\) and \(e^{ax}\). By breaking down the product into parts, you aim to make the integration process more manageable.
The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]By choosing the appropriate \(u\) and \(dv\), you can break a complex integral into more concise pieces.
The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]By choosing the appropriate \(u\) and \(dv\), you can break a complex integral into more concise pieces.
- Choose \(u\) to be a function that becomes simpler when differentiated (here, \(x^n\)),
- Choose \(dv\) to be a part easy to integrate (such as \(e^{ax} \, dx\)),
- Differentiate \(u\) to find \(du\).
- Integrate \(dv\) to find \(v\).
Indefinite Integrals
Indefinite integrals represent the family of all antiderivatives of a function, expressed without specific limits. The aim is to find a function whose derivative replicates the initial integrand, thus capturing all possible solutions.
When using methods like integration by parts and reduction formulas, you’re often dealing with indefinite integrals because the overall goal is to find a general form of the antiderivative. For integration by parts, for example, one may repeatedly apply it in the context of indefinite integrals:
For the expression \(\int x^2 e^{-3x} \, dx\), employing integration by parts and reduction was utilized to transform and simplify the integral into a form where standard integration could then be applied. This recursive and methodical simplification is at the heart of dealing with indefinite integrals, paving the way for efficiently finding your antiderivative.
The process often reflects the strategic breakdown of complex functions until they align with recognizable forms, spotlighting the elegance and utility of indefinite integrals in calculus.
When using methods like integration by parts and reduction formulas, you’re often dealing with indefinite integrals because the overall goal is to find a general form of the antiderivative. For integration by parts, for example, one may repeatedly apply it in the context of indefinite integrals:
For the expression \(\int x^2 e^{-3x} \, dx\), employing integration by parts and reduction was utilized to transform and simplify the integral into a form where standard integration could then be applied. This recursive and methodical simplification is at the heart of dealing with indefinite integrals, paving the way for efficiently finding your antiderivative.
The process often reflects the strategic breakdown of complex functions until they align with recognizable forms, spotlighting the elegance and utility of indefinite integrals in calculus.
Other exercises in this chapter
Problem 36
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(a) Show that $$ \frac{1}{\sqrt{1+x^{2}}} \geq \frac{1}{2 x}>0 $$ for \(x \geq 1\). (b) Use your result in (a) to show that $$ \int_{1}^{\infty} \frac{1}{\sqrt{
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