Problem 30
Question
17-40. Evaluate each improper integral or state that it is divergent. $$ \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} d x $$
Step-by-Step Solution
Verified Answer
The integral is divergent.
1Step 1: Identify the Type of Improper Integral
The given integral is \( \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} \, dx \). It is an improper integral because the upper limit is infinity. We deal with improper integrals by using limits.
2Step 2: Set Up the Limit Definition
Replace the upper limit with a variable, say \( b \), and then take the limit as \( b \to \infty \). Thus, we have:\[ \lim_{b \to \infty} \int_{0}^{b} \frac{x^{2}}{x^{3}+1} \, dx \]
3Step 3: Use a Substitution Method
Let \( u = x^{3} + 1 \), which implies \( du = 3x^{2} \, dx \) or \( x^{2} \, dx = \frac{du}{3} \). Change the limits of integration as \( x = 0 \) implies \( u = 1 \) and \( x = b \) implies \( u = b^3 + 1 \). Thus the integral becomes:\[ \frac{1}{3} \lim_{b \to \infty} \int_{1}^{b^3+1} \frac{1}{u} \, du \]
4Step 4: Integrate Using the New Variable
Integrate \( \frac{1}{u} \) with respect to \( u \), which results in:\[ \frac{1}{3} \lim_{b \to \infty} [ \ln|u| ]_{1}^{b^3+1} \]
5Step 5: Evaluate the Definite Integral
Substitute the new limits into the antiderivative:\[ \frac{1}{3} \lim_{b \to \infty} [ \ln(b^3+1) - \ln(1) ] \]Since \( \ln(1) = 0 \), we have:\[ \frac{1}{3} \lim_{b \to \infty} \ln(b^3+1) \]
6Step 6: Analyze the Limit
As \( b \to \infty \), \( \ln(b^3+1) \to \infty \) because the logarithmic function increases unboundedly when its argument goes to infinity. Therefore, the limit is infinity.
7Step 7: Conclusion on Convergence
Since \( \lim_{b \to \infty} \ln(b^3+1) = \infty \), the original integral does not converge. It is divergent.
Key Concepts
Limits in CalculusDivergent IntegralsSubstitution Method
Limits in Calculus
In calculus, limits are a fundamental tool, especially when it comes to evaluating improper integrals. They provide a method for handling infinity, which cannot be used directly in calculations.
Imagine a situation where we have an integral with infinity as one of its limits, such as the one given in this problem. Here’s where limits step in: they help us make sense of the indefinite behavior occurring at infinity by setting up a variable to represent the "boundaries" of our integral, which then extends to infinity.
This strategy involves substituting infinity with a variable (say, \( b \)), and then exploring how the integral behaves as this variable approaches infinity. By setting up a limit, we can analyze the integral’s behavior, determining whether it converges to a finite value or diverges, leading to an endless value.
In our example, the limit we set is \( \lim_{b \to \infty} \int_{0}^{b} \frac{x^{2}}{x^{3}+1} \, dx \). This setup allows us to apply calculus concepts to understand and solve the integral.
Imagine a situation where we have an integral with infinity as one of its limits, such as the one given in this problem. Here’s where limits step in: they help us make sense of the indefinite behavior occurring at infinity by setting up a variable to represent the "boundaries" of our integral, which then extends to infinity.
This strategy involves substituting infinity with a variable (say, \( b \)), and then exploring how the integral behaves as this variable approaches infinity. By setting up a limit, we can analyze the integral’s behavior, determining whether it converges to a finite value or diverges, leading to an endless value.
In our example, the limit we set is \( \lim_{b \to \infty} \int_{0}^{b} \frac{x^{2}}{x^{3}+1} \, dx \). This setup allows us to apply calculus concepts to understand and solve the integral.
Divergent Integrals
An integral is considered divergent when it does not converge to a specific finite value. In simpler terms, as you extend the "bounds" out indefinitely, like in our improper integral here, the outcome tends to infinity instead of stabilizing to a number.
In the problem at hand, after setting up and evaluating the limits, it is determined that the integral \( \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} \, dx \) is divergent.
Why does this happen? The core lies in the logarithmic function \( \ln(b^3 + 1) \), which was part of our integration result. As \( b \to \infty \), \( \ln(b^3+1) \to \infty \), so we observe an unbounded increase.
This essentially confirms that the integral’s result does not settle to a particular number but instead continues on to infinity. Understanding divergent integrals helps in grasiving the behavior of complex systems that grow unrestrained, a useful insight in advanced calculus.
In the problem at hand, after setting up and evaluating the limits, it is determined that the integral \( \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} \, dx \) is divergent.
Why does this happen? The core lies in the logarithmic function \( \ln(b^3 + 1) \), which was part of our integration result. As \( b \to \infty \), \( \ln(b^3+1) \to \infty \), so we observe an unbounded increase.
This essentially confirms that the integral’s result does not settle to a particular number but instead continues on to infinity. Understanding divergent integrals helps in grasiving the behavior of complex systems that grow unrestrained, a useful insight in advanced calculus.
Substitution Method
The substitution method is a crucial technique in calculus for simplifying integrals to make them easier to solve.
When you encounter a complex integral, like \( \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} \, dx \), substitution is your friendly tool that can transform it into a more manageable problem.
In our scenario, a clever choice of substitution was made: by letting \( u = x^3 + 1 \), the messy algebraic expression is simplified, transforming the integration problem into a straightforward form \( \int \frac{1}{u} \, du \). This step involves more than just changing variables; it requires adjusting the limits too.
Originally, these limits of integration were from \( x = 0 \) to \( \infty \). Post-substitution, they changed to \( u = 1 \) and \( u = b^3 + 1 \).
This method doesn’t just make the math tidier. It opens the road for further analytical strategies, even letting us use integration techniques suited to simple logs, as seen in our step-by-step solution. Mastering substitution in calculus makes tackling complex problems much less daunting.
When you encounter a complex integral, like \( \int_{0}^{\infty} \frac{x^{2}}{x^{3}+1} \, dx \), substitution is your friendly tool that can transform it into a more manageable problem.
In our scenario, a clever choice of substitution was made: by letting \( u = x^3 + 1 \), the messy algebraic expression is simplified, transforming the integration problem into a straightforward form \( \int \frac{1}{u} \, du \). This step involves more than just changing variables; it requires adjusting the limits too.
Originally, these limits of integration were from \( x = 0 \) to \( \infty \). Post-substitution, they changed to \( u = 1 \) and \( u = b^3 + 1 \).
This method doesn’t just make the math tidier. It opens the road for further analytical strategies, even letting us use integration techniques suited to simple logs, as seen in our step-by-step solution. Mastering substitution in calculus makes tackling complex problems much less daunting.
Other exercises in this chapter
Problem 29
Use integration by parts to find each integral. \(\int x e^{a x} d x \quad(a \neq 0)\)
View solution Problem 29
Find each integral by using the integral table on the inside back cover. $$ \int \frac{1}{\sqrt{4-e^{2 t}}} d t $$
View solution Problem 30
Use integration by parts to find each integral. \(\int(x+b) e^{a x} d x \quad(a \neq 0)\)
View solution Problem 30
Find each integral by using the integral table on the inside back cover. $$ \int \frac{e^{t}}{9-e^{2 t}} d t $$
View solution