Chapter 1: Problem 17
In the following exercises, find each indefinite integral by using appropriate substitutions. $$ \int x e^{-x^{2}} d x $$
Short Answer
Expert verified
The indefinite integral is \(-\frac{1}{2} e^{-x^2} + C\).
Step by step solution
01
Identify the substitution variable
We need to choose a substitution that simplifies the integral. Notice that the exponent in the function suggests using substitution. Let's set \( u = -x^2 \). This works because the derivative \( du = -2x \, dx \) closely resembles part of the integrand \( x \, dx \).
02
Solve for dx in terms of u
From \( u = -x^2 \), we have that \( du = -2x \, dx \). Solving for \( x \, dx \), we get \( x \, dx = -\frac{1}{2} du \).
03
Change the integral in terms of u
Substitute \( x \, dx = -\frac{1}{2} du \) and \( e^{-x^2} = e^{u} \) into the integral. The integral becomes: \[ \int x e^{-x^2} \, dx = \int -\frac{1}{2} e^{u} \, du \].
04
Integrate with respect to u
The integral \( \int -\frac{1}{2} e^{u} \, du \) can be integrated as \( -\frac{1}{2} \int e^{u} \, du \). The integral of \( e^{u} \) is \( e^{u} \), so we have: \[ -\frac{1}{2} e^{u} + C \].
05
Substitute back in terms of x
Recall that \( u = -x^2 \). Substitute back to get the expression in terms of \( x \): \[ -\frac{1}{2} e^{-x^2} + C \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful tool for solving integrals, especially when direct integration is challenging. By selecting an appropriate substitution, we can transform a complex integrand into a simpler one. In the given problem, the aim is to simplify \( \int x e^{-x^{2}} \, dx \).To start, look for a part of the integrand that complicates the integral. Often, this is something inside a function like an exponent or a square root. Here, we notice that the exponent \(-x^2\) suggests a clever substitution of \( u = -x^2 \).Once a substitution is chosen, find its derivative. Here, \( du = -2x \, dx \). Our goal is to replace \( x \, dx \) with an expression in terms of \( u \). Solve for \( x \, dx \): \( x \, dx = -\frac{1}{2} du \).This simplification allows us to rewrite the integral in terms of \( u \), making it much easier to handle. This method is essential for tackling a wide variety of integration challenges.
Integration Techniques
Learning different integration techniques is crucial when solving a wide range of indefinite integrals. The substitution method is just one of these techniques, but it's often used when it's not straightforward to integrate directly.When you apply substitution to our integral \( \int x e^{-x^{2}} \, dx \), we translate the original integral into a simpler form. After substitution, it becomes:\[ \int -\frac{1}{2} e^{u} \, du \].This is a much simpler integral to evaluate — it's now a basic integral of an exponential function. Ignoring the constant factor \(-\frac{1}{2}\), the integral of \( e^{u} \) is straightforward to compute since it remains \( e^{u} \).Using substitution, turning the problem into an integration involving basic forms simplifies the problem. The familiarity with these techniques not only gives a structured path to follow but also speeds up the solving process significantly.
Exponential Functions
Exponential functions often intimidate students at first due to their rapid growth properties and their unique behavior. However, when it comes to integration, exponential functions like \( e^x \) have a special property: their derivative is the same as the function itself. This makes its anti-derivative direct and simple.In the given exercise, the function \( e^{-x^2} \) only differs slightly. Although the exponent makes it non-standard, using substitution simplifies it significantly. After substitution, we integrate \( e^u \), which directly returns itself after integration, plus any constant factor:\[ -\frac{1}{2} e^{u} + C. \]Ultimately, the exponential function's behavior, combined with the method of substitution, allows us to solve seemingly complex integrals with ease. Understanding these special features of exponentials is crucial in calculus. They frequently appear in solutions, showcasing their natural occurrence in mathematical problems across various topics.