Finding The Greatest Value Of An Integral Function F(x)

The problem at hand seeks to determine the greatest value of the function F(x), which is defined as the definite integral of the absolute value of t from 1 to x, where x lies within the closed interval [-1/2, 1/2]. This problem combines concepts from real analysis, specifically integration, definite integrals, and the determination of maxima and minima. The absolute value function introduces a piecewise nature to the integrand, requiring careful consideration when evaluating the integral. To solve this, we need to analyze how the integral behaves over the given interval and identify the point where F(x) attains its maximum value. This involves understanding the properties of definite integrals, the behavior of the absolute value function, and how to apply these concepts to find the maximum value of a function on a closed interval. The Leibniz integral rule, although mentioned, isn't strictly necessary for this particular problem, as we can directly evaluate the integral by considering the piecewise definition of the absolute value function. The key lies in breaking the integral into appropriate intervals where the absolute value can be expressed without the absolute value signs, thereby allowing for straightforward integration. The analysis of the resulting function will then lead us to the determination of its maximum value within the specified domain.

To begin, let's delve deeper into the function F(x), defined as the integral of the absolute value of t from 1 to x: F(x) = ∫[1 to x] |t| dt. The core of this function lies in the absolute value function, |t|, which behaves differently depending on the sign of t. Specifically, |t| = t when t ≥ 0, and |t| = -t when t < 0. This piecewise nature of the absolute value function is crucial for correctly evaluating the definite integral. Given the interval of interest for x, which is [-1/2, 1/2], and the lower limit of integration being 1, we observe that the interval of integration spans both positive and potentially negative values of t. However, since x is always less than 1, the direction of integration is from 1 to a value less than 1. This means we need to consider two cases to properly evaluate the integral. When x is positive (0 ≤ x ≤ 1/2), the integration is over positive values of t, and |t| simply equals t. However, when x is negative (-1/2 ≤ x < 0), the integration still involves only positive values of t because the interval of integration is [1, x], and even the smallest value of x (-1/2) is greater than 0. Therefore, throughout the interval [-1/2, 1/2], we can treat |t| as t. This simplifies the integral evaluation significantly. We can now express F(x) as the definite integral of t from 1 to x, which is a much simpler integral to compute. Understanding this nuanced behavior of the absolute value function within the context of the definite integral is the key to unlocking the solution to this problem. The next step is to perform the integration and then analyze the resulting function to find its maximum value within the given domain. This analysis will involve calculus techniques, such as finding critical points and evaluating the function at the endpoints of the interval.

Now, let's proceed with the evaluation of the definite integral that defines F(x). As established in the previous section, since we are integrating |t| from 1 to x and considering the domain x ∈ [-1/2, 1/2], we can replace |t| with t. This simplification is valid because even when x is negative, the integration is performed from 1 to x, which implies that we are traversing from a larger positive value to a smaller value, but still within the realm of positive values of t. Therefore, the integral becomes: F(x) = ∫[1 to x] t dt. This is a standard integral that can be easily evaluated using the power rule for integration. The power rule states that ∫t^n dt = (t^(n+1))/(n+1) + C, where C is the constant of integration. In our case, n = 1, so the integral of t is (t^2)/2. Applying this, we get: F(x) = [(x^2)/2] evaluated from 1 to x. To evaluate the definite integral, we substitute the limits of integration: F(x) = (x^2)/2 - (1^2)/2. This simplifies to: F(x) = (x^2)/2 - 1/2. Now we have a simple quadratic function representing F(x). This quadratic function is a parabola opening upwards, which means it has a minimum value, but within the context of our interval [-1/2, 1/2], we are interested in finding its maximum value. The next step involves analyzing this quadratic function over the given interval to determine where the greatest value occurs. This will likely involve considering the endpoints of the interval and any critical points within the interval. The ease with which we were able to evaluate this integral highlights the importance of carefully considering the properties of the integrand and the interval of integration before diving into complex calculations. Recognizing that |t| could be simplified to t over the relevant range was a critical step in making the problem tractable.

Having computed F(x) = (x^2)/2 - 1/2, our next crucial step is to determine the maximum value of this function within the interval x ∈ [-1/2, 1/2]. Since F(x) is a quadratic function represented by a parabola opening upwards, its vertex corresponds to the minimum value. However, the maximum value within a closed interval will occur at one of the endpoints of the interval, unless there's a critical point within the interval that dictates otherwise. To find the critical points, we would typically take the derivative of F(x) and set it to zero. Let's compute the derivative of F(x) with respect to x: F'(x) = d/dx [(x^2)/2 - 1/2] = x. Setting F'(x) = 0, we find that x = 0 is the only critical point. Now, we need to evaluate F(x) at the critical point and the endpoints of the interval to determine the maximum value. The endpoints are x = -1/2 and x = 1/2. Let's calculate F(x) at these points:

• At x = -1/2: F(-1/2) = ((-1/2)^2)/2 - 1/2 = (1/4)/2 - 1/2 = 1/8 - 1/2 = -3/8
• At x = 1/2: F(1/2) = ((1/2)^2)/2 - 1/2 = (1/4)/2 - 1/2 = 1/8 - 1/2 = -3/8
• At x = 0: F(0) = ((0)^2)/2 - 1/2 = -1/2

Comparing these values, we see that F(-1/2) = F(1/2) = -3/8, and F(0) = -1/2. Since -3/8 is greater than -1/2, the maximum value of F(x) on the interval [-1/2, 1/2] is -3/8. This result indicates that the greatest value of the integral occurs at both endpoints of the interval. The fact that the critical point yields a lower value further confirms that the maximum is indeed at the boundaries. This comprehensive analysis, involving the evaluation of the integral, finding critical points, and comparing values at endpoints, leads us to the final solution.

In conclusion, we have successfully determined the greatest value of the function F(x) = ∫[1 to x] |t| dt for x in the interval [-1/2, 1/2]. By carefully analyzing the absolute value function, evaluating the definite integral, and considering the behavior of the resulting quadratic function, we found that the maximum value of F(x) is -3/8. This maximum occurs at both endpoints of the interval, x = -1/2 and x = 1/2. The problem highlights the importance of understanding the properties of definite integrals and absolute value functions, as well as the techniques for finding maxima and minima of functions on closed intervals. The initial simplification of the integral, by recognizing that |t| could be replaced with t over the relevant range, was a key step in making the problem tractable. Furthermore, the analysis of the quadratic function, including finding its critical point and evaluating it at the endpoints, provided a robust method for determining the maximum value. This exercise serves as a valuable example of how concepts from real analysis and calculus can be applied to solve problems involving integrals and function optimization. The systematic approach, from understanding the function to evaluating the integral and finally finding the maximum value, demonstrates a clear and logical problem-solving process that can be applied to similar challenges in mathematics.