Problem 17 - Applied Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} \, dx \).

Step 3: Compute the Indefinite Integral

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

Step 4: Apply the Fundamental Theorem of Calculus

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

Step 5: 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 \( \frac{1}{2b^2} \to 0 \).

Step 6: Conclusion

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

Key Concepts

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.