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. $$\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \sqrt{4-c_{k}^{2}} \Delta x, \quad[0,1]$$

Short answer

The limit of the given Riemann sum as \(n \rightarrow \infty\) is the definite integral of the function \(\sqrt{4-c^2}\) over the interval [0,1], which can be written as \(\int_{0}^{1} \sqrt{4-c^2} \,dc\).

Step-by-step solution

Recognize the Problem Type

The given expression is the limit of the sum as \(n \rightarrow \infty\) of \(\sqrt{4-c_{k}^{2}} \Delta x\), where each \(c_{k}\) is chosen from the \(k\)th subinterval of a partition of the interval [0, 1] into \(n\) subintervals each of length \(\Delta x\). This kind of structure represents a Riemann sum, which can be transformed into a definite integral as \(n \rightarrow \infty\).

Identify the Function and Interval

The function being considered here is \(\sqrt{4-c^{2}}\), and the interval over which this function is being summed is [0, 1]. The variable \(c\) refers to a point picked within each subinterval when the interval [0, 1] is divided into \(n\) parts. As \(n \rightarrow \infty\), the width \(\Delta x\) of the subintervals approaches 0 and \(c\) becomes any point in the interval [0, 1].

Transform Riemann Sum into Definite Integral

Since the Riemann sum in this exercise represents the estimate of the area under the curve of function \(f(c) = \sqrt{4-c^2}\) from 0 to 1, when \(n \rightarrow \infty\), the sum turns into the definite integral of \(f(c)\) over the interval [0, 1]. This transformation is based on the definition of the definite integral as the limit of a Riemann sum. Hence, the Riemann sum can be written as: \(\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \sqrt{4-c_{k}^{2}} \Delta x = \int_{0}^{1} \sqrt{4-c^2} \,dc\).