Problem 10 - Essential Calculus

Question

In Exercises 9-14, write the limit as a definite integral on the interval \([a, b],\) where \(c_{i}\) is any point in the \(i\) th subinterval. $$ \frac{\text { Limit }}{\lim _{\|\Delta\| \rightarrow 0} \sum_{i=1}^{n} 6 c_{i}\left(4-c_{i}\right)^{2} \Delta x_{i}} \quad \frac{\text { Interval }}{[0,4]} $$

Step-by-Step Solution

Answer

The limit of the given Riemann sum is equivalent to the definite integral \(\int_{0}^{4} 6x\left(4-x\right)^{2} dx\).

Step 1: Recognizing the Limit of the Riemann Sum

The limit of the Riemann sum in the problem is given by: \[ \lim _{\|\Delta\| \rightarrow 0} \sum_{i=1}^{n} 6 c_{i} \left(4-c_{i}\right)^{2} \Delta x_{i} \]This is a definition of the definite integral, where the function \(f(x)\) is given by \(6x(4-x)^2\). The \(\Delta x_{i}\) term signals the limit of partition widths going to zero, which shows that we are dealing with a definite integral. In a Riemann sum, \(c_{i}\) represents a selected point in the i-th subinterval, in a definite integral, it is generally represented by \(x\). So we can replace \(c_{i}\) with \(x\).

Step 2: Writing the Limit as a Definite Integral

The limit of the Riemann sum is equivalent to a definite integral over the interval [0,4]. The function \(f(x)\) is the term in the sum that depends on \(c_{i}\), or in other words, \(f(x) = 6x(4-x)^2\). Hence, the equivalent definite integral is:\[\int_{0}^{4} 6x\left(4-x\right)^{2} dx\]This represents the same quantity as the Riemann sum in the limit as the maximum size of the subintervals goes to zero.

Key Concepts

Riemann Sum

A Riemann sum is a method for approximating the total area under a curve on a graph, commonly used as an approach to approximate the value of a definite integral. It involves splitting the area into many thin vertical slices and finding the sum of the areas of these slices. In simple terms, you are breaking down the curve into many small rectangles.

The process begins by partitioning the interval \([a, b]\) into smaller subintervals. Each of these subintervals has a width of \(\Delta x_i\), which is calculated as the difference between the endpoints of the subinterval. Then, a sample point \(c_i\) from each subinterval is selected, at which the function will be evaluated.

The basic form of a Riemann sum is \( \sum_{i=1}^{n} f(c_i) \Delta x_i \), where \(f(c_i)\) is the value of the function at the chosen points. As the number of partitions increases (the subintervals become thinner), the Riemann sum gets closer to the actual value of the definite integral. This sum becomes exact as the maximum width of the subintervals approaches zero, leading to our next concept, which is the limit of partition widths.

Limit of Partition Widths

The 'limit of partition widths' refers to the process of making the partitions or subintervals infinitely thin, effectively increasing their number to infinity. In other words, each subinterval width \(\Delta x_i\) approaches zero. This concept ensures that the Riemann sum converges to the definite integral's precise value.

By allowing \(|\Delta|\) to approach zero, we make sure that we have a finer and finer approximation of the area under the curve, eventually leading to a true "area" representation under the curve, rather than just an estimation. This refinement process is significant since it transforms a Riemann sum into a definite integral, defined precisely over an interval \([a, b]\). A complete understanding of this concept is fundamental for grasping the inherent precision and elegance of integral calculus.

Importantly, this involves taking the limit: \( \lim _{\|\Delta\| \rightarrow 0} \sum_{i=1}^{n} f(c_i) \Delta x_i \). As the partitions get smaller, this sum approaches the true value of the definite integral, which brings us to the importance of function integration.

Function Integration

Function integration is a core concept in calculus, dealing with finding the integral of a function over a specific interval. When you integrate a function, you're essentially accumulating the area under the curve defined by that function, within the bounds given. This is represented as a definite integral, which provides a number that represents the total accumulation over that range.

This process is closely tied to the previously discussed concepts. By transitioning from a Riemann sum, with its limit of partition widths going to zero, we obtain the definite integral. For our specific exercise, the function to be integrated, \(f(x) = 6x(4-x)^2\), is integrated over the interval \[0, 4\].

The integral is calculated as follows: \( \int_{0}^{4} 6x(4-x)^2 \, dx \). This yields the exact area under the curve of the function between the limits of 0 and 4. Function integration thus converts the sum of infinitely small changes (from the integral calculus perspective) into an overall result, which gives profound insights into the behavior of functions over specific intervals.