Problem 61 - Calculus Early Transcendentals

Question

Determine whether the following statements are true and give an explanation or counterexample. a. If \(f\) is continuous and \(0p\) d. If \(\int_{1}^{\infty} x^{-p} d x\) exists, then \(\int_{1}^{\infty} x^{-q} d x\) exists, where \(q>p\) e. \(\int_{1}^{\infty} \frac{d x}{x^{3 p+2}}\) exists, for \(p>-\frac{1}{3}\)

Step-by-Step Solution

Answer

Based on the given step-by-step solution, answer the following question: Question: Determine if the following statement is true or false: "If the integral \(\int_{1}^{\infty} x^{-p} dx\) converges for some real number p, then the integral \(\int_{1}^{\infty} x^{-q} dx\) also converges for any real number q > p." Answer: The statement is true. If the integral \(\int_{1}^{\infty} x^{-p} dx\) converges, then \(1p\), \(1

Step 1: a. Statement A: Analysis

In order to determine if the integral exists, we will apply the Comparison theorem for improper integrals. Given that \(0

Step 2: a. Statement A: Conclusion

The statement is true. By the Comparison theorem for improper integrals, if \(0 \leq f(x) \leq g(x)\) on the interval \([0, \infty)\) and the integral of g(x) converges, then the integral of f(x) will also converge. Therefore, \(\int_{0}^{\infty} f(x) dx\) exists.

Step 3: b. Statement B: Analysis

We know that \(\lim_{x \rightarrow \infty}f(x) = 1\). To determine if the integral exists, we will consider the behavior of the function f(x) in the interval \([0,\pi]\).

Step 4: b. Statement B: Conclusion

The statement is true. As f(x) has a finite limit at infinity, and the integral is over a bounded interval (\([0, \pi]\)), the definite integral \(\int_{0}^{\pi} f(x) dx\) exists.

Step 5: c. Statement C: Analysis

This statement deals with improper integrals of the form \(\int_{0}^{1} x^{-p} dx\) and \(\int_{0}^{1} x^{-q} dx\). We will check under what conditions each integral converges.

Step 6: c. Statement C: Conclusion

The statement is false. For the integral \(\int_{0}^{1} x^{-p} dx\) to converge, we need \(-1p\), it is not guaranteed that the second integral will converge. For example, consider \(p=0\) and \(q=2\): \(\int_{0}^{1} x^{0} dx\) converges, but \(\int_{0}^{1} x^{-2} dx\) does not.

Step 7: d. Statement D: Analysis

This statement is quite similar to statement C, but the integral is now over the interval \([1, \infty)\). We will check the conditions for the convergence of integrals of the form \(\int_{1}^{\infty} x^{-p} dx\) and \(\int_{1}^{\infty} x^{-q} dx\).

Step 8: d. Statement D: Conclusion

The statement is true. If the integral \(\int_{1}^{\infty} x^{-p} dx\) converges, then \(1p\), \(1

Step 9: e. Statement E: Analysis

The statement deals with the convergence of the integral \(\int_{1}^{\infty} \frac{dx}{x^{3p+2}}\). We need to determine whether the integral converges for \(p > -\frac{1}{3}\).

Step 10: e. Statement E: Conclusion

The statement is true. For the integral \(\int_{1}^{\infty} \frac{dx}{x^{3p+2}}\) to converge, we need \(1<3p+2<\infty\), which simplifies to \(p > -\frac{1}{3}\). Since \(p > -\frac{1}{3}\), the integral exists.

Key Concepts

Comparison Theorem for Improper Integrals

Understanding the behavior of functions as they approach infinity is crucial when dealing with improper integrals. The Comparison Theorem for Improper Integrals is a tool that helps us determine the convergence of such integrals. Essentially, it states that if you have two functions, f(x) and g(x), such that 0 \<= f(x) \<= g(x) for all x in the interval of interest, and you know that the integral of g(x) converges, then the integral of f(x) also converges.

Let's apply this theorem to a given exercise. If we have f(x) and g(x) where both are continuous and 0 < f(x) < g(x) on [0, \infty), and we know that \(\int_{0}^{\infty} g(x) dx = M < \infty\), the integral of f(x) also exists. This is because f(x) is always less than g(x), which converges; hence, f(x)'s integral is bounded above by M.

For easier understanding, think of it as being on a road trip and following a car that is guaranteed to reach its destination. If your car never goes faster and always stays behind, you're bound to reach the destination as well, assuming there are no roadblocks specific to your path.

Convergence of Improper Integrals

Determining the convergence of improper integrals involves checking whether the integral of a function, typically as x approaches infinity or a point of discontinuity, has a finite value. Convergence is not always straightforward to assess as it can depend on the behavior of the function over the interval. For instance, consider an improper integral where the limit of f(x) as x approaches infinity is known.

If \(\lim_{x \rightarrow \infty}f(x) = 1\), we might infer the integral converges over a finite interval, such as [0, \pi], because the function approaches a finite limit and the interval does not extend to infinity. In the exercise provided, this means that \(\int_{0}^{\pi} f(x) dx\) exists. However, this is not a general rule, and each case must be examined individually. The convergence can differ significantly when dealing with intervals that approach zero or infinity, which often require different techniques or theorems to determine.

Furthermore, for an integral over the interval [1, \infty) and the function x^{-p}, the integral converges if p is greater than one. This is because as x grows larger, the function shrinks rapidly enough to produce a finite area under the curve — like stretching a rubber band until it can stretch no further, indicating a limit to the potential 'stretch' or, in our case, integral value.

Definite Integral Existence

The existence of a definite integral can be analyzed using the bounds of integration and the behavior of the function within those bounds. A definite integral exists if the function is continuous over a closed and bounded interval, such as [a, b]. Issues arise when the function is not continuous over the interval or the interval is unbounded — that is, it includes infinity.

In the context of the examples from the exercise, for an integral of the form \(\int_{1}^{\infty} \frac{dx}{x^{3p+2}}\) to exist, the powers must align so that the function decays quickly enough as x approaches infinity. For example, when p > -\frac{1}{3}, the denominator grows rapidly enough to ensure that the total area under the curve from 1 to \infty remains finite – imagine gradually adding smaller and smaller slices to a pie, eventually adding an infinite number of slices that don't substantially increase the size of the pie. If the function is bounded and converges to a point as x approaches infinity, its tail may be long but it doesn’t have enough 'weight' to make the integral diverge — ensuring the existence of the integral over the specified interval.

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Key Concepts

Comparison Theorem for Improper Integrals

Understanding the behavior of functions as they approach infinity is crucial when dealing with improper integrals. The Comparison Theorem for Improper Integrals is a tool that helps us determine the convergence of such integrals. Essentially, it states that if you have two functions, f(x) and g(x), such that 0 \<= f(x) \<= g(x) for all x in the interval of interest, and you know that the integral of g(x) converges, then the integral of f(x) also converges.

Let's apply this theorem to a given exercise. If we have f(x) and g(x) where both are continuous and 0 < f(x) < g(x) on [0, \infty), and we know that \(\int_{0}^{\infty} g(x) dx = M < \infty\), the integral of f(x) also exists. This is because f(x) is always less than g(x), which converges; hence, f(x)'s integral is bounded above by M.

For easier understanding, think of it as being on a road trip and following a car that is guaranteed to reach its destination. If your car never goes faster and always stays behind, you're bound to reach the destination as well, assuming there are no roadblocks specific to your path.

Convergence of Improper Integrals

Determining the convergence of improper integrals involves checking whether the integral of a function, typically as x approaches infinity or a point of discontinuity, has a finite value. Convergence is not always straightforward to assess as it can depend on the behavior of the function over the interval. For instance, consider an improper integral where the limit of f(x) as x approaches infinity is known.

If \(\lim_{x \rightarrow \infty}f(x) = 1\), we might infer the integral converges over a finite interval, such as [0, \pi], because the function approaches a finite limit and the interval does not extend to infinity. In the exercise provided, this means that \(\int_{0}^{\pi} f(x) dx\) exists. However, this is not a general rule, and each case must be examined individually. The convergence can differ significantly when dealing with intervals that approach zero or infinity, which often require different techniques or theorems to determine.

Furthermore, for an integral over the interval [1, \infty) and the function x^{-p}, the integral converges if p is greater than one. This is because as x grows larger, the function shrinks rapidly enough to produce a finite area under the curve — like stretching a rubber band until it can stretch no further, indicating a limit to the potential 'stretch' or, in our case, integral value.

Definite Integral Existence

The existence of a definite integral can be analyzed using the bounds of integration and the behavior of the function within those bounds. A definite integral exists if the function is continuous over a closed and bounded interval, such as [a, b]. Issues arise when the function is not continuous over the interval or the interval is unbounded — that is, it includes infinity.

In the context of the examples from the exercise, for an integral of the form \(\int_{1}^{\infty} \frac{dx}{x^{3p+2}}\) to exist, the powers must align so that the function decays quickly enough as x approaches infinity. For example, when p > -\frac{1}{3}, the denominator grows rapidly enough to ensure that the total area under the curve from 1 to \infty remains finite – imagine gradually adding smaller and smaller slices to a pie, eventually adding an infinite number of slices that don't substantially increase the size of the pie. If the function is bounded and converges to a point as x approaches infinity, its tail may be long but it doesn’t have enough 'weight' to make the integral diverge — ensuring the existence of the integral over the specified interval.