/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 17-40. Evaluate each improper in... [FREE SOLUTION] | 91Ó°ÊÓ

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17-40. Evaluate each improper integral or state that it is divergent. $$ \int_{1}^{\infty} \frac{1}{x^{3}} d x $$

Short Answer

Expert verified
The integral converges to \( \frac{1}{2} \).

Step by step solution

01

Identify Type of Improper Integral

The integral \( \int_{1}^{\infty} \frac{1}{x^3} \, dx \) is an improper integral because its upper limit is infinity.
02

Set Up the Limit for Evaluation

To evaluate the improper integral, we replace the infinity symbol with a variable \( b \) and then take the limit as \( b \to \infty \). Thus, the integral becomes \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} \, dx \).
03

Compute the Indefinite Integral

Find the antiderivative of \( \frac{1}{x^3} \). The antiderivative is \( -\frac{1}{2x^2} \).
04

Apply the Fundamental Theorem of Calculus

Using the antiderivative from the previous step, evaluate the definite integral from 1 to \( b \): \( \left[ -\frac{1}{2x^2} \right]_1^b = -\frac{1}{2b^2} + \frac{1}{2} \).
05

Take the Limit as b Approaches Infinity

Take the limit of the expression as \( b \to \infty \): \( \lim_{b \to \infty} \left(-\frac{1}{2b^2} + \frac{1}{2}\right) = \frac{1}{2} \), since \( \frac{1}{2b^2} \to 0 \).
06

Conclusion

The limit exists and equals \( \frac{1}{2} \), hence the integral converges to \( \frac{1}{2} \).

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Key Concepts

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

Convergence
When evaluating improper integrals, the concept of convergence is crucial. Convergence means that as we evaluate the integral over an infinite interval, it approaches a specific value. In our exercise, the improper integral is given as \( \int_{1}^{\infty} \frac{1}{x^3} \, dx \). The main goal is to discover whether this integral converges to an actual number or diverges to infinity.To determine convergence, we set up the integral with a finite limit \( b \) and take the limit as \( b \to \infty \). If the resulting value exists and is finite, the integral converges. In this case, by substituting the limits and solving, we found it converges to \( \frac{1}{2} \).
  • Convergent integral: Result approaches a finite number.
  • Divergent integral: Result does not settle at a finite value.
Understanding convergence helps in determining whether an improper integral produces meaningful results.
Antiderivatives
Antiderivatives play a central role in calculating integrals. An antiderivative of a function is another function whose derivative yields the original function. For the integrand \( \frac{1}{x^3} \), its antiderivative is \( -\frac{1}{2x^2} \). In simpler terms, we're looking for a function that, when differentiated, returns \( \frac{1}{x^3} \).Finding antiderivatives allows us to evaluate definite integrals, especially through the Fundamental Theorem of Calculus, which connects derivatives, antiderivatives, and definite integrals.
  • Antiderivative of \( \frac{1}{x^n} \): \( \frac{-1}{(n-1)x^{n-1}} \) for \( n eq 1 \).
  • Utilization: Crucial in transitioning an indefinite integral to a definite one.
Thus, mastering antiderivatives is essential for solving a wide range of calculus problems.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a powerful tool that links the concept of antiderivatives with definite integrals. It provides a systematic way to evaluate an integral once its antiderivative is known.In our problem, once the antiderivative \( -\frac{1}{2x^2} \) of \( \frac{1}{x^3} \) is found, the theorem helps us evaluate the integral from the lower limit 1 to an upper limit \( b \). We use the expression:\[ \left[ -\frac{1}{2x^2} \right]_1^b = -\frac{1}{2b^2} + \frac{1}{2} \]
  • Step-by-step application: Substitute the limits into the antiderivative.
  • Simplifies calculations: Converts complex differential to an arithmetic problem.
Using the theorem not only simplifies computation but also confirms the integral's convergence to the value \( \frac{1}{2} \), providing a complete analysis of the integral's behavior.

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