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Magazine Chapter 6: Problem 15
Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{1}^{\infty} x e^{-3 x^{2}} d x $$
Short Answer
Expert verified
The integral converges and evaluates to \( \frac{1}{6e^{3}} \).
Step by step solution
01
Understanding the Problem
The given integral is \( \int_{1}^{\infty} x e^{-3 x^{2}} \, dx \). It's an improper integral because it has an infinite upper limit. Our goal is to determine whether this integral converges (has a finite value) or diverges.
02
Set Up the Limit Definition of the Improper Integral
We express the improper integral as a limit: \[\lim_{{b \to \infty}} \int_{1}^{b} x e^{-3x^{2}} \, dx\]This representation allows us to deal with the infinite upper limit by considering the integral as \( b \) approaches infinity.
03
Use Substitution to Simplify the Integral
To evaluate the integral, use the substitution \( u = -3x^{2} \). Then, \( du = -6x \, dx \), so \( dx = \frac{du}{-6x} \).Hence, the integral becomes:\[\int x e^{-3x^{2}} \, dx = \int e^{u} \cdot \frac{du}{-6} \]
04
Integrate in Terms of \( u \)
Substituting into the integral and integrating with respect to \( u \), we get: \[\int e^{u} \cdot \frac{du}{-6} = -\frac{1}{6} \int e^{u} \, du = -\frac{1}{6} e^{u} + C\]Return to the original variable: \[-\frac{1}{6} e^{-3x^{2}} + C\]
05
Evaluate the Definite Integral with Limits
Return the limits of integration back to \( x \) and evaluate the definite integral:\[\int_{1}^{b} x e^{-3x^{2}} \, dx = \left[ -\frac{1}{6} e^{-3x^{2}} \right]_{1}^{b}\]Apply the limits to get:\[\left( -\frac{1}{6} e^{-3b^{2}} \right) - \left( -\frac{1}{6} e^{-3}\right)\]
06
Determine the Limit as \( b \to \infty \)
Evaluate the limit:\[\lim_{{b \to \infty}} \left( -\frac{1}{6} e^{-3b^{2}} + \frac{1}{6} e^{-3} \right)\]As \( b \to \infty \), \( e^{-3b^{2}} \to 0 \), so the limit becomes:\[\frac{1}{6} e^{-3}\]Therefore, the integral converges to \( \frac{1}{6e^{3}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a technique used to simplify integrals, making them easier to calculate. By substituting part of an integral with a new variable, we can transform it into a simpler form.
For the given integral \( \int x e^{-3x^{2}} \, dx \), using substitution helps us handle the exponential part. Here's how it works:
For the given integral \( \int x e^{-3x^{2}} \, dx \), using substitution helps us handle the exponential part. Here's how it works:
- Choose a new variable \( u \). In this case, let \( u = -3x^{2} \).
- Find the differential of \( u \), which is \( du = -6x \, dx \).
- Rearrange the equation for \( dx \), giving \( dx = \frac{du}{-6x} \).
Convergence and Divergence
In calculus, convergence or divergence refers to whether an improper integral results in a finite value or not.
An improper integral like \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \) has an infinite limit, which requires us to determine if it converges or diverges. We test convergence by expressing the integral as a limit:
An improper integral like \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \) has an infinite limit, which requires us to determine if it converges or diverges. We test convergence by expressing the integral as a limit:
- Write \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \) as \( \lim_{{b \to \infty}} \int_{1}^{b} x e^{-3x^{2}} \, dx \).
- After substitution and integration, apply the limits to find the behavior as \( b \to \infty \).
Definite Integral
A definite integral computes the accumulated area between a function and the x-axis over a given interval. It has upper and lower limits that define this range.
While calculating the integral \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \), we handled it using limits in calculus. We treated \( \infty \) with a variable substitution and ultimately evaluated it over \([1, b]\), where \( b \to \infty \).
While calculating the integral \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \), we handled it using limits in calculus. We treated \( \infty \) with a variable substitution and ultimately evaluated it over \([1, b]\), where \( b \to \infty \).
- The transformed integral becomes \( -\frac{1}{6} e^{-3x^{2}} \) evaluated from 1 to \( b \).
- Substitute back \( x \) and integrate to apply limits.
Limits in Calculus
Limits in calculus help us understand the behavior of functions as they approach a certain point or value. This is especially crucial for improper integrals which involve infinity.
For the integral \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \), employing limits is vital:
For the integral \( \int_{1}^{\infty} x e^{-3x^{2}} \, dx \), employing limits is vital:
- We transform the integral \( \int_{1}^{b} x e^{-3x^{2}} \, dx \) considering \( b \to \infty \).
- The idea is to determine what happens to the integral's value as \( b \) gets larger and larger.
- As \( b \to \infty \), \( e^{-3b^{2}} \to 0 \), simplifying the limit to a finite result: \( \frac{1}{6e^{3}} \).
