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Evaluate the integrals using integration by parts. $$ \int x^{2} e^{-x} d x $$

Short Answer

Expert verified
The integral \( \int x^2 e^{-x} \, dx \) evaluates to \(-e^{-x} (x^2 + 2x + 2) + C\).

Step by step solution

01

Identify Parts for Integration by Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). We need to choose \( u \) and \( dv \) such that the integration becomes easier when applying this rule. For \( \int x^2 e^{-x} \, dx \), we choose \( u = x^2 \) and \( dv = e^{-x} \, dx \).
02

Differentiate and Integrate Chosen Parts

Differentiate \( u = x^2 \) to find \( du \). This gives us \( du = 2x \, dx \). Next, find \( v \) by integrating \( dv = e^{-x} \, dx \), which gives \( v = -e^{-x} \).
03

Apply the Integration by Parts Formula

Substitute the expressions for \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \[ \int x^2 e^{-x} \, dx = x^2 (-e^{-x}) - \int (-e^{-x}) \cdot 2x \, dx. \] Simplifying, we get: \[ \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 \int x e^{-x} \, dx. \]
04

Repeat Integration by Parts for Remaining Integral

Now, we need to evaluate \( \int x e^{-x} \, dx \) using integration by parts again. Choose \( u = x \) and \( dv = e^{-x} \, dx \). Thus, \( du = dx \) and \( v = -e^{-x} \). Applying the formula again: \[ \int x e^{-x} \, dx = x(-e^{-x}) - \int (-e^{-x}) \, dx. \] Simplify this to get: \[ \int x e^{-x} \, dx = -xe^{-x} - e^{-x}. \]
05

Substitute Back into the Original Expression

Substitute \( \int x e^{-x} \, dx = -xe^{-x} - e^{-x} \) back into our expression from Step 3: \[ \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2(-xe^{-x} - e^{-x}). \] Simplify the expression to: \[ \int x^2 e^{-x} \, dx = -x^2 e^{-x} - 2xe^{-x} - 2e^{-x}. \]
06

Express the Final Answer

Combine the terms and factor out the common \( e^{-x} \) term resulting in: \[ \int x^2 e^{-x} \, dx = -e^{-x} (x^2 + 2x + 2) + C, \] where \( C \) is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integrals
Indefinite integrals are a fundamental concept in calculus. They represent the family of functions whose derivative is the given function. This concept is essential because it allows us to find the antiderivative or the original function before differentiation.
When working with indefinite integrals, we're often provided an expression like \(\int f(x) \, dx\). Our task is to determine a function \(F(x)\) such that its derivative \(F'(x)\) equals \(f(x)\). Because integration is the reverse process of differentiation, we say that \(F(x)\) is the antiderivative.
For indefinite integrals, there's a vital component called the constant of integration, denoted as \(C\). Since differentiation eliminates constant terms, when we find the antiderivative, we add this constant term to include all possible solutions.
  • Represents family of antiderivatives
  • Form: \(F(x) + C\)
  • Helps recover original functions from derivatives
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They exhibit rapid growth or decay, which makes them crucial in many fields.
A typical form of an exponential function is \(b^x\), where \(b\) is the base. In calculus, the exponential function \(e^x\) is exquisitely important due to its unique property: the derivative of \(e^x\) is itself, \(e^x\).
In our problem, we dealt with \(e^{-x}\), which represents exponential decay. The derivative of \(e^{-x}\) is \(-e^{-x}\), demonstrating how exponential functions maintain simplicity in differentiation. A clear understanding of these properties assists in performing operations such as integration by parts efficiently.
  • Form: \(b^x\)
  • Natural base: \(e\), easier differentiation/integration
  • Key in growth/decay models
Polynomial Functions
Polynomial functions consist of terms made up of variables raised to whole number exponents and constant coefficients. They're some of the most common functions you'll encounter in calculus.
Such functions can take the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\). In the exercise, we have a specific polynomial term \(x^2\), which we chose as our \(u\) in integration by parts.
The derivative of \(x^2\) is simple to compute, resulting in \(2x\). This straightforward differentiation is why polynomial functions are often chosen as the \(u\) term in integration by parts. These functions help simplify complex integration tasks and split the problem into manageable parts.
  • Characterized by variable powers and coefficients
  • Essential for modeling a wide variety of phenomena
  • Simplicity in differentiation and integration
Integration Techniques
Integration techniques, such as integration by parts, greatly simplify the process of finding antiderivatives for complex functions.
Integration by parts is particularly useful when we're dealing with a product of functions, such as a polynomial function and an exponential function. The formula is \( \int u \, dv = uv - \int v \, du \).
The key to using this method effectively is choosing the right functions for \(u\) and \(dv\). Typically, \(u\) is selected to be a polynomial, because differentiating polynomials reduces their degree. On the other hand, \(dv\) should be the more complex function, like the exponential part, since they are often simpler to integrate.
When applied, integration by parts reduces the complexity of an integral, allowing you to solve otherwise challenging problems step by step.
  • Useful for products of functions
  • Formula: \(\int u \, dv = uv - \int v \, du\)
  • Choice of \(u\) and \(dv\) affects simplicity

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