Line Integral Calculation: Direct Substitution Method

Hey everyone! Let's dive into calculating line integrals using the direct substitution method. This approach can be super handy when you have a clear relationship between your variables, making the integration process a bit smoother. In this article, we're going to break down a specific example step by step, so you can see exactly how it works. We'll start by understanding the problem, then walk through the substitution, the integration itself, and finally discuss the result. So, grab your calculators and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we understand the problem. We are given a line integral ∮_C F · dr, where we need to evaluate this integral over a specific curve C. The curve C is defined by the equation y = 2x², and the variable x ranges from 0 to 1. This gives us a clear path in the xy-plane to follow. The vector field F is implicitly defined within the integral expression, and our goal is to compute the line integral by directly substituting the relationship between x and y provided by the curve C. This method is particularly useful when the curve can be expressed explicitly as a function of a single variable, which simplifies the integral significantly. By understanding the problem setup, we can appreciate the elegance of the direct substitution method and its effectiveness in solving line integrals.

The key here is to recognize that we're not just dealing with a regular integral; we're dealing with a line integral. This means we're integrating along a path, not just an interval on the x-axis. The path is defined by y = 2x², which gives us a clear relationship between x and y. This relationship is crucial because it allows us to express everything in terms of a single variable, x, making the integral solvable. We're given that x goes from 0 to 1, which provides the limits of integration. The expression ∮_C F · dr represents the line integral itself, where F is a vector field, and dr is an infinitesimal displacement vector along the curve C. Our mission, should we choose to accept it (and we do!), is to compute this integral using the direct substitution method. This involves plugging in the equation of the curve into the integral and evaluating it over the given limits. So, let's roll up our sleeves and get into the nitty-gritty of the calculation!

Direct Substitution and Setup

Alright, let's get to the heart of the matter: direct substitution. This is where we take the given relationship between x and y (y = 2x²) and plug it directly into the line integral. This is a crucial step because it transforms the line integral, which is inherently two-dimensional (or even three-dimensional in some cases), into a regular integral with respect to a single variable (x in this case). The beauty of this method lies in its ability to simplify complex integrals by leveraging the specific path we're integrating along. We're essentially parameterizing the curve, expressing it in terms of a single variable, which makes the integration process much more manageable. Think of it like converting a winding mountain path into a straight line on a map – we're simplifying the journey to make it easier to navigate. Now, let's see how this works in practice with our specific example.

In our problem, we have the line integral ∫_C F · dr, and we're told that y = 2x². We're also given the expression inside the integral: [-3x² dx + 5x(2x²) d(2x²)]. Notice how this expression already hints at the substitution we need to make. We have terms involving x, dx, and d(2x²), which perfectly align with our given relationship y = 2x². The dx term represents the infinitesimal change in x, and the d(2x²) term represents the infinitesimal change in y, expressed in terms of x. This is exactly what we need to perform the substitution. We're essentially replacing dy with its equivalent expression in terms of x and dx. This step is critical because it allows us to write the entire integral in terms of a single variable, x, which is something we know how to handle. So, we're taking a multi-dimensional problem and reducing it to a one-dimensional problem, which is a powerful simplification technique in calculus.

Performing the Integration

Now comes the fun part: performing the integration. We've set up the integral by directly substituting y = 2x², and we've got an expression that's entirely in terms of x. This means we can now use our standard integration techniques to find the value of the integral. Remember, integration is essentially the reverse process of differentiation. We're looking for a function whose derivative matches the expression inside the integral. This can sometimes be a bit of a puzzle, but with practice, you'll start to recognize common patterns and techniques that make the process smoother. In our case, the expression is a polynomial, which makes it relatively straightforward to integrate. We'll use the power rule for integration, which states that the integral of x^n is (x^(n+1))/(n+1), plus a constant of integration. However, since we're dealing with a definite integral (with limits of integration), we won't need to worry about the constant of integration, as it will cancel out when we evaluate the integral at the upper and lower limits.

Our integral, after substitution, looks like this: ∫_0^1 (-3x² + 5x(2x²) d(2x²)). Let's simplify this further. First, we need to find d(2x²). Using the power rule for differentiation, we get d(2x²) = 4x dx. Substituting this back into the integral, we have: ∫_0^1 [-3x² dx + 5x(2x²)(4x dx)] = ∫_0^1 [-3x² dx + 40x^4 dx]. Now, we can combine the terms inside the integral: ∫_0^1 (-3x² + 40x^4) dx. This is a polynomial expression, which is much easier to integrate. We apply the power rule to each term separately: The integral of -3x² is -x³, and the integral of 40x^4 is 8x^5. So, the antiderivative is -x³ + 8x^5. Now, we need to evaluate this antiderivative at the limits of integration, which are 0 and 1. This means we'll plug in x = 1 and x = 0 into the antiderivative and subtract the results. This will give us the final value of the definite integral.

Evaluating the Definite Integral

Okay, we've reached the final stretch: evaluating the definite integral. We've found the antiderivative, which is -x³ + 8x^5, and now we need to plug in the limits of integration and subtract. This step is crucial because it gives us the numerical value of the line integral. Remember, the definite integral represents the signed area under the curve (or, in the case of a line integral, the