Evaluating The Definite Integral Of 3x^5 From 2 To 4

Introduction to Definite Integrals

In the realm of calculus, definite integrals play a pivotal role in determining the accumulated effect of a function over a specified interval. They are a fundamental concept with wide-ranging applications across various fields, including physics, engineering, economics, and statistics. The definite integral, denoted by the symbol ∫ with upper and lower limits, quantifies the signed area between a function's graph and the x-axis within the given interval. This area represents the net change of the function's antiderivative over that interval. Understanding definite integrals is crucial for solving problems related to accumulation, such as calculating displacement from velocity, total cost from marginal cost, or probability from probability density functions.

To truly grasp the essence of definite integrals, it's essential to distinguish them from their counterparts, indefinite integrals. While indefinite integrals yield a family of functions representing all possible antiderivatives of a given function, definite integrals produce a single numerical value. This value precisely captures the net area under the curve between the specified limits of integration. The process of evaluating a definite integral involves finding the antiderivative of the integrand, evaluating it at the upper and lower limits, and then subtracting the value at the lower limit from the value at the upper limit. This procedure, formalized by the Fundamental Theorem of Calculus, provides a powerful tool for calculating areas and solving accumulation problems.

The applications of definite integrals extend far beyond the theoretical realm of mathematics. In physics, they are used to compute work done by a force, the center of mass of an object, and the total charge of a distributed charge density. Engineers utilize definite integrals to design structures, analyze circuits, and model fluid flow. Economists employ them to calculate consumer surplus, producer surplus, and present values of future income streams. Statisticians rely on definite integrals to determine probabilities and expected values of continuous random variables. The versatility of definite integrals stems from their ability to quantify accumulation, making them an indispensable tool in any field dealing with rates of change and total quantities.

Problem Statement: ∫24 3x5 dx

In this article, we will delve into the process of evaluating the definite integral ∫24 3x5 dx, a quintessential example that showcases the application of the power rule for integration and the Fundamental Theorem of Calculus. This particular integral represents the signed area under the curve of the function f(x) = 3x5 between the limits x = 2 and x = 4. The integrand, 3x5, is a polynomial function, making it amenable to the power rule, a cornerstone technique in integral calculus. This rule states that the integral of xn, where n is any real number except -1, is (xn+1)/(n+1) plus a constant of integration. The constant of integration is crucial for indefinite integrals, but it cancels out when evaluating definite integrals, as we will see later.

The limits of integration, 2 and 4, define the interval over which we are calculating the accumulated effect of the function. The lower limit, 2, marks the starting point, while the upper limit, 4, marks the ending point. These limits are crucial for determining the specific numerical value of the definite integral. They essentially bound the area we are interested in calculating. Without these limits, we would only be finding the general antiderivative, which represents a family of functions rather than a single value. The definite integral, with its specified limits, provides a precise measure of the net change of the antiderivative over the given interval.

Before we embark on the actual evaluation, it's beneficial to have a conceptual understanding of what the integral represents. The function f(x) = 3x5 is a monotonically increasing function for positive values of x. This means that as x increases, the value of f(x) also increases. In the interval [2, 4], the function will be above the x-axis, so the definite integral will represent a positive area. Visualizing the graph of the function can provide valuable intuition about the expected result. The area under the curve will be substantial, given the steepness of the function and the width of the interval. This preliminary analysis sets the stage for a more confident and insightful evaluation of the integral.

Step-by-Step Solution

To effectively evaluate the definite integral ∫24 3x5 dx, we will meticulously follow a step-by-step approach, ensuring clarity and accuracy in each stage. This methodical process not only yields the correct answer but also reinforces a deeper understanding of the underlying principles of integral calculus. The steps involved are as follows:

1. Find the Antiderivative

The initial step in evaluating a definite integral is to determine the antiderivative of the integrand. In this case, our integrand is 3x5. To find its antiderivative, we employ the power rule for integration, which states that ∫xn dx = (xn+1)/(n+1) + C, where C is the constant of integration. Applying this rule to our integrand, we first recognize the constant multiple of 3. According to the constant multiple rule of integration, we can pull this constant outside the integral sign. Thus, we have 3∫x5 dx. Now, applying the power rule to x5, we add 1 to the exponent and divide by the new exponent, resulting in (x6)/6. Multiplying this by the constant 3, we obtain 3(x6)/6, which simplifies to (x6)/2. Therefore, the antiderivative of 3x5 is (x6)/2 + C. It's crucial to remember the constant of integration, C, when finding indefinite integrals. However, as we will see in the next step, this constant cancels out when evaluating definite integrals.

2. Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus provides the cornerstone for evaluating definite integrals. It states that if F(x) is an antiderivative of f(x), then ∫ab f(x) dx = F(b) - F(a). In simpler terms, to evaluate a definite integral, we find the antiderivative of the integrand, evaluate it at the upper limit of integration, evaluate it at the lower limit of integration, and then subtract the value at the lower limit from the value at the upper limit. This theorem elegantly connects the concepts of differentiation and integration, providing a powerful tool for solving a wide range of problems. In our case, we have already found the antiderivative of 3x5 to be (x6)/2 + C. Now, we need to evaluate this antiderivative at the upper limit, 4, and the lower limit, 2, and then subtract the results.

3. Evaluate at the Upper and Lower Limits

Now we evaluate the antiderivative, (x6)/2 + C, at the upper limit, x = 4, and the lower limit, x = 2. At the upper limit, we have (46)/2 + C = (4096)/2 + C = 2048 + C. At the lower limit, we have (26)/2 + C = (64)/2 + C = 32 + C. Notice that the constant of integration, C, is present in both evaluations. This is a crucial observation because, as we will see in the next step, this constant will cancel out when we subtract the two values. This cancellation is why we often omit the constant of integration when evaluating definite integrals. The evaluation at the limits provides us with the numerical values of the antiderivative at the endpoints of the interval, which are essential for determining the net change.

4. Subtract and Simplify

The final step in evaluating the definite integral is to subtract the value of the antiderivative at the lower limit from its value at the upper limit. This step directly applies the Fundamental Theorem of Calculus. We have F(4) - F(2) = (2048 + C) - (32 + C). Distributing the negative sign, we get 2048 + C - 32 - C. Notice that the constants of integration, C, cancel each other out, leaving us with 2048 - 32. Performing the subtraction, we obtain 2016. Therefore, the definite integral ∫24 3x5 dx evaluates to 2016. This numerical value represents the signed area under the curve of the function 3x5 between the limits x = 2 and x = 4. The positive value indicates that the area is entirely above the x-axis, as expected given the behavior of the function in this interval.

Final Answer

Therefore, after meticulously following the steps of finding the antiderivative, applying the Fundamental Theorem of Calculus, evaluating at the upper and lower limits, and simplifying, we arrive at the final answer:

∫24 3x5 dx = 2016

This result represents the exact value of the definite integral, quantifying the signed area under the curve of the function 3x5 between the limits x = 2 and x = 4. The positive value of 2016 confirms our initial expectation that the area lies entirely above the x-axis. This comprehensive solution demonstrates the power and precision of integral calculus in determining accumulated effects and areas under curves. The process highlights the importance of understanding the Fundamental Theorem of Calculus and the application of integration rules. By breaking down the problem into manageable steps, we have not only obtained the correct answer but also gained a deeper appreciation for the underlying concepts and techniques involved in evaluating definite integrals. This knowledge serves as a solid foundation for tackling more complex integration problems in the future.