Problem 25 - Calculus for Biology and Medicine

Question

Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{0}^{2} \frac{1}{(x-1)^{4}} d x $$

Step-by-Step Solution

Answer

The integral is divergent because it tends to \(-\infty\).

Step 1: Identify the Problematic Point

The function \( f(x) = \frac{1}{(x-1)^4} \) becomes undefined if \( x = 1 \). Therefore, this is a potential point of discontinuity, making the integral improper at \( x = 1 \). Since the integration range includes \( x = 1 \), evaluate the limit as \( x \) approaches 1.

Step 2: Set Up the Limit for Convergence Analysis

To analyze the convergence, we split the integral at the discontinuity. Consider two limits: \( \lim_{a \to 1^-} \int_0^a \frac{1}{(x-1)^4} \, dx \) and \( \lim_{b \to 1^+} \int_b^2 \frac{1}{(x-1)^4} \, dx \). Evaluate these limits separately to determine whether the integral converges.

Step 3: Evaluate Left-Hand Limit

Evaluate \( \lim_{a \to 1^-} \int_0^a \frac{1}{(x-1)^4} \, dx \). The antiderivative of \( \frac{1}{(x-1)^4} \) is \( -\frac{1}{3(x-1)^3} \), thus:\[ \lim_{a \to 1^-} \left[ -\frac{1}{3(x-1)^3} \right]_0^a = \lim_{a \to 1^-} \left( -\frac{1}{3(a-1)^3} + \frac{1}{3} \right) \]As \( a \to 1^- \), \(\frac{1}{(a-1)^3} \to +\infty\), so the limit is \(-\infty\).

Step 4: Evaluate Right-Hand Limit

Evaluate \( \lim_{b \to 1^+} \int_b^2 \frac{1}{(x-1)^4} \, dx \). Using the antiderivative, calculate:\[ \lim_{b \to 1^+} \left[ -\frac{1}{3(x-1)^3} \right]_b^2 = \lim_{b \to 1^+} \left(-\frac{1}{3(b-1)^3} + \frac{1}{3} \right) \]As \( b \to 1^+ \), \(\frac{1}{(b-1)^3} \to +\infty\), so this limit also results in \(-\infty\).

Step 5: Conclude the Integral's Convergence

Since both left-hand and right-hand limits tend towards \(-\infty\), the integral \( \int_0^2 \frac{1}{(x-1)^4} \, dx \) does not converge. An integral diverges if any part around the discontinuity tends to infinity or does not result in a finite number.

Key Concepts

Improper Integrals

Improper integrals are integrals with at least one of two main issues: the limits of integration are infinite, or the integrand becomes infinite within the interval of integration. In our example, the improper integral is formed because the integrand \( \frac{1}{(x-1)^4} \) becomes infinite at \( x = 1 \). This occurs within the interval \([0, 2]\), given that the denominator approaches zero at \( x = 1 \), making the function undefined and dramatically large as it nears this point.

• Improper integrals often require splitting into limits to handle points of discontinuity.
• These integrals need careful evaluation to determine if they converge (result in a finite value) or diverge (do not result in a finite value).

In essence, improper integrals require a limit process to determine their behavior over the range of integration where standard calculus rules don't apply neatly. Understanding when and why an integral is considered improper is crucial for successful calculus problem-solving.

Limits in Calculus

Limits play an essential role in calculus, particularly when dealing with improper integrals. A limit helps describe the behavior of a function as the input approaches a certain point. With our improper integral, we effectively used limits to assess the behavior near the point of discontinuity, \( x = 1 \). To handle this, we evaluated the limits from both the left and right of the discontinuity.

• In the context of improper integrals, limits tell us how the function behaves as it gets closer to its domain's problematic points.
• Evaluating these limits helps establish whether an integral converges or diverges.

In our example, the lack of a finite limit on either side of \( x = 1 \) directly led to concluding that the integral diverges. Thus, understanding limits and how to evaluate them is imperative for resolving improper integral problems effectively.

Discontinuity and Integration

Discontinuity refers to points where a function is not smoothly defined. When integrating, any discontinuity usually results in an improper integral. In the given exercise, \( x = 1 \) is a discontinuity since \( \frac{1}{(x-1)^4}\) becomes infinite there. Such discontinuities must be carefully managed using limits to assess their effect on the overall integral.

• Discontinuities can make a simple integral complex due to undefined behavior at certain points.
• Integration with discontinuities involves breaking down the problem using limits to analyze each segmented part.

In our discussion, because both sides of the discontinuity approached infinity, the function diverged when nearing \( x = 1 \) from both directions. Recognizing and handling discontinuities using well-structured limit evaluations is integral in calculus when improper integrals are involved.