Definite Integral: Solving ∫ 1/(2x-1)^2 Dx From 0 To 2

Let's dive into evaluating the definite integral: ∫[0, 2] 1/(2x-1)^2 dx. This problem involves a bit of calculus, specifically the application of definite integrals. We'll break it down step-by-step, making sure we handle any potential pitfalls along the way. So, buckle up, folks, and let's get started!

Understanding the Integral

Before we jump into the calculations, let's understand what we're dealing with. We're asked to find the definite integral of the function f(x) = 1/(2x-1)^2 within the limits of integration from 0 to 2. Essentially, we are finding the area under the curve of this function between these two points on the x-axis. Now, it is important to notice that the function has a potential discontinuity where the denominator equals zero. This is something we absolutely must consider when solving.

When faced with a definite integral, it's crucial to first examine the integrand, which is the function we're integrating (in this case, 1/(2x-1)^2). A keen eye will spot that the denominator, (2x-1)^2, can equal zero. This occurs when 2x - 1 = 0, which simplifies to x = 1/2. This is a critical point because it lies within our interval of integration [0, 2]. The presence of this discontinuity means we can't directly apply the Fundamental Theorem of Calculus without some extra care. Remember, the Fundamental Theorem of Calculus requires the function to be continuous over the interval of integration. Since our function is discontinuous at x = 1/2, we'll need to split the integral into two parts, one approaching 1/2 from the left and one approaching from the right. This ensures we handle the discontinuity appropriately and avoid arriving at an incorrect answer. This initial assessment is a vital step in tackling definite integrals, especially those that might harbor discontinuities. Recognizing these points early on can save us from major headaches and lead us to the correct solution more efficiently.

Breaking Down the Integral

Because of the discontinuity at x = 1/2, we need to split our integral into two parts:

∫[0, 2] 1/(2x-1)^2 dx = ∫[0, 1/2] 1/(2x-1)^2 dx + ∫[1/2, 2] 1/(2x-1)^2 dx

Now, we'll evaluate each of these integrals separately. The key here is to treat them as improper integrals, using limits to approach the point of discontinuity. This is a standard technique when dealing with integrals that have discontinuities within their interval of integration. By using limits, we can carefully examine the behavior of the function as it gets closer and closer to the problematic point, without actually plugging the discontinuity value into the function. This allows us to obtain a meaningful result, even when the function is not defined at a particular point. The process involves replacing the discontinuity with a variable (like 't') and taking the limit as 't' approaches the discontinuity. This gives us a precise way to evaluate the integral and determine if it converges to a finite value or diverges. It's like tiptoeing around the edge of a cliff – we can get as close as we want, but we never actually fall off. This method ensures we adhere to the rigorous rules of calculus and arrive at a correct solution.

Evaluating the First Integral

Let's tackle the first integral: ∫[0, 1/2] 1/(2x-1)^2 dx. We'll rewrite it using a limit:

lim (t→1/2-) ∫[0, t] 1/(2x-1)^2 dx

To solve this, we first find the indefinite integral of 1/(2x-1)^2. Using a simple u-substitution (u = 2x - 1, du = 2 dx), we get:

∫ 1/(2x-1)^2 dx = -1/2 * (1/(2x-1)) + C

Now, we apply the limits of integration (0 and t) and then take the limit as t approaches 1/2 from the left. This step is critical because it allows us to handle the discontinuity at x = 1/2 carefully. Instead of directly substituting 1/2 into the integral, which would lead to an undefined expression, we use a limit to see what happens as we get arbitrarily close to 1/2. This technique transforms the integral into a form that we can actually evaluate. We substitute the upper limit (t) and the lower limit (0) into the antiderivative, subtract the results, and then evaluate the limit as t approaches 1/2 from the left. This process reveals how the function behaves near the discontinuity and whether the integral converges to a finite value or diverges. It's a powerful tool for dealing with integrals that have singularities or other problematic points within their interval of integration.

Evaluating the Second Integral

Next, we evaluate the second integral: ∫[1/2, 2] 1/(2x-1)^2 dx. We do a similar process, using a limit:

lim (t→1/2+) ∫[t, 2] 1/(2x-1)^2 dx

Using the same indefinite integral we found earlier, we apply the limits of integration (t and 2) and then take the limit as t approaches 1/2 from the right. This step mirrors the process we used for the first integral, but it approaches the discontinuity from the opposite direction. This is crucial because the behavior of the function can differ significantly as we approach a discontinuity from the left versus from the right. By evaluating the limit as t approaches 1/2 from the right, we gain a complete picture of the function's behavior around the singularity. We substitute the upper limit (2) and the lower limit (t) into the antiderivative, subtract the results, and then evaluate the limit. If the limit exists and is finite, it means the integral converges; if the limit is infinite or does not exist, the integral diverges. This careful consideration of the limit from both sides of the discontinuity ensures we accurately evaluate the definite integral.

Putting It All Together

After evaluating both limits, we'll find that both integrals diverge. This means that the original integral also diverges. Divergence in the context of integrals implies that the area under the curve is unbounded, essentially stretching out to infinity. This can occur when a function has a vertical asymptote within the interval of integration, as is the case here. When we split the integral around the discontinuity and evaluated the limits, we found that the limits themselves went to infinity. This is a clear indicator that the area under the curve is not finite, and therefore, the integral diverges. It's a crucial concept in calculus, highlighting that not all integrals have a finite value. Understanding divergence helps us interpret the behavior of functions and the areas they enclose, providing a more comprehensive understanding of integral calculus. In practical terms, it tells us that there is no single numerical answer to the integral; instead, it represents an unbounded quantity.

Conclusion

The original integral ∫[0, 2] 1/(2x-1)^2 dx diverges due to the discontinuity at x = 1/2. Therefore, none of the provided answer choices (A. 1.5, B. 2/3, C. -2/3, D. 2.5, E. 3) are correct. When dealing with definite integrals, it's crucial to check for discontinuities within the interval of integration. If a discontinuity exists, you'll likely need to split the integral and use limits to correctly evaluate it. Remember, guys, calculus is all about precision and attention to detail!