Question

In Exercises \(1-6,\) each \(c_{k}\) is chosen from the \(k\) th subinterval of a regular partition of the indicated interval into \(n\) subintervals of length \(\Delta x .\) Express the limit as a definite integral. $$\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{1-c_{k}} \Delta x, \quad[2,3]$$

Short answer

The expression can be simplified as \(\int_{2}^{3} \frac{1}{1-x} dx\). This is the definite integral form of the given limit of a series.

Step-by-step solution

Identify the function

First recognize that the function to be integrated is \(f(c_k)= \frac{1}{1-c_{k}}\). This step is important for understanding how the summation translates into an integral.

Identify the interval

The interval of integration is indicated as \([2,3]\). This interval corresponds to the each \(c_{k}\) chosen from the \(k\) th subinterval of a regular partition of the interval [2,3].

Identify the variable of integration

The variable of integration (which replaces \(c_k\) in the limit of the series) is usually \(x\), hence, \(f(x)= \frac{1}{1-x}\).

Write the definite integral

Replace the limit of the summation with the definite integral over the interval [2,3]. Each \(c_{k}\) in the function after summing is replaced by the variable of integration, \(x\). Thus the expression becomes \(\int_{2}^{3} \frac{1}{1-x} dx\).

Key concepts

Riemann Sums

Understanding the concept of Riemann sums is essential in the world of integral calculus. A Riemann sum is a method of approximating the area under a curve by breaking it into small 'slices' or rectangles. When we calculate the area of each slice and add them up, we are essentially performing a Riemann sum.

Imagine you are trying to find out how much paint you would need to cover a particular shape on a piece of paper; if you cover the shape with tiny squares and add up the area of those squares, you're employing the same idea as a Riemann sum.

In the exercise given, Riemann sums come into play as we divide the interval from 2 to 3 into subintervals of length \(\Delta x\) and sum the function's values at specified points within these subintervals, multiplied by \(\Delta x\). As the number of subintervals \(n\) increases, the width of each rectangle \(\Delta x\) gets smaller and our Riemann sum becomes a better approximation of the exact area under the curve.

Limit of a Summation

The limit of a summation is a fundamental concept when dealing with infinite series. In calculus, we often want to find the exact value of a sum when the number of terms is infinitely large. The task is to determine what value the summation approaches as the number of terms grows without bound.

In our given problem, we are asked to take the limit as \(n\) approaches infinity, which means we are looking at the area under the curve as the sum of an infinite number of infinitesimal slices. This process transforms our Riemann sum from an approximate value to the exact value of the definite integral. It's like counting infinite grains of sand one by one—the limit tells us how much sand we have when we count endlessly.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total amount, or 'aggregate,' under curves on graphs. This involves calculating areas, volumes, and other concepts that arise from summing an infinite number of infinitesimally small factors.

Imagine you are at the beach and want to build a sandcastle. You have to gather sand from all over the beach—a little from here, a little from there—until you have enough to build your castle. Integral calculus does something similar; it gathers the 'bits and pieces' of area under a curve to find the whole.

In the context of our problem, integral calculus is used to transition from the sum of finite rectangles in a Riemann sum to a precise area under the curve, represented by the definite integral \(\int_{2}^{3} \frac{1}{1-x} dx\). This move from discrete to continuous is what integral calculus is all about—finding the whole by considering the sum of its parts.

Interval of Integration

The interval of integration is the range on the x-axis over which we are integrating. Simply put, it tells us the starting and ending points of the area we are interested in finding on a graph.

For instance, if we were to travel from one city to another, the interval of integration would be like the stretch of road we need to cover on our journey. It defines the bounds within which we sum up all those little slices we talked about with Riemann sums and limits of summations.

In the exercise, the interval of integration is \[2,3\], meaning we are looking at the area under the function \(f(x)= \frac{1}{1-x}\) from \(x=2\) to \(x=3\). By computing the definite integral over this interval, we find the exact area or total amount that the function accumulates between these two points on the x-axis.