Problem 13 - 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. $$ \frac{\text { Limit }}{\lim _{\|\Delta\| \rightarrow 0} \sum_{i=1}^{n}\left(1+\frac{3}{c_{i}}\right) \Delta x_{i}} \quad \frac{\text { Interval }}{[1,5]} $$

Step-by-Step Solution

Answer

The limit as a definite integral on the interval \([1,5]\) is \(\int_{1}^{5}(1+\frac{3}{x})dx\).

Step 1: Identifying the function

First, recognize the function inside the sum in the limit. Here \((1+\frac{3}{c_{i}})\) can be rewritten as \(1+\frac{3}{x}\), as \(c_{i}\) represents any point within the \(i^{th}\) subinterval, and can be generalized as 'x'.

Step 2: Identifying the interval

The interval over which we are taking the limit is provided as \([1,5]\). This means 'a' is 1 and 'b' is 5.

Step 3: Writing the limit as a definite integral

Finally, the limit of the Riemann sum can be written as a definite integral over the interval \([a,b]\) for the function \(1+\frac{3}{x}\). This gives us the required definite integral as: \(\int_{1}^{5}(1+\frac{3}{x})dx\).

Key Concepts

Riemann Sum

The concept of a Riemann sum is fundamental in understanding how integrals work. Named after the mathematician Bernhard Riemann, the Riemann sum is a method of approximating the total area under a curve—often referred to as the integral of a function—over a certain interval. To do this, the interval \( [a, b] \) is divided into smaller subintervals, and for each subinterval, a rectangle is created. The height of this rectangle is the value of the function at a chosen point within the subinterval, often denoted as \( c_i \). By summing the areas of these rectangles, we obtain an approximation of the area under the curve.

The more subintervals we have, the better our approximation becomes. Therefore, as we let the maximum width of these subintervals, \( \|\Delta\| \), approach zero, our Riemann sum becomes a better and better approximation of the actual integral. This is what's happening in the exercise when setting up the limit: as \( \|\Delta\| \to 0 \), the sum \( \sum_{i=1}^{n}\left(1+\frac{3}{c_{i}}\right) \Delta x_{i} \) gets closer to the exact area under the function \( 1+\frac{3}{x} \) over the interval \( [1,5] \).

Limit of a Function

The limit of a function is a core concept in calculus, which describes the behavior of a function as the input approaches a certain value. In the context of integrals and Riemann sums, we are interested in the limit as the maximum width of subintervals goes to zero (\( \|\Delta\| \to 0 \) ). This process effectively smooths out the approximation given by a Riemann sum into a precise measurement—the definite integral.

Visualize this as the transition from a staircase-like graph of rectangles (the Riemann sum) to a smooth curve (the actual function graph). The key idea is that although individual Riemann sums are only approximations, the limit governs the transition to an exact value. With regard to the exercise, \( \lim_{\|\Delta\| \to 0} \sum_{i=1}^{n}\left(1+\frac{3}{c_{i}}\right) \Delta x_{i} \) represents the exact area under the curve from \( x = 1 \) to \( x = 5 \) for the function \( 1+\frac{3}{x} \) as the width of the rectangles in the Riemann sum vanishes.

Integration

Integration is the grand concept that wraps up the discussion of areas under curves and Riemann sums. It is essentially a formal way to determine the area under a curve and is symbolized by the integral sign ( \( \int \) ). Through the process of integration, which is the inverse operation to differentiation, we can calculate the accumulation of quantities, such as area, volume, and other concepts that add up infinitesimal pieces.

In practical applications, integration allows us to solve a variety of problems in physics, engineering, and economics, among many other fields. In the context of the exercise, we expressed the limit of a Riemann sum as a definite integral. That is, the area under \( 1+\frac{3}{x} \) from \( x=1 \) to \( x=5 \) is given by \( \int_{1}^{5}(1+\frac{3}{x})dx \). This is a powerful statement, as we have moved from an approximation to an exact value for the area. In solving the definite integral, we apply techniques from calculus to find this precise area, which is in many cases a feasible and incredibly useful task.