Problem 39 - Calculus Single Variable

Question

Suppose that \(f(x) \geq 0\) for \(x\) in \(I=[a, b]\). If for each subinterval \(\left[x_{j-1}, x_{j}\right]\) that arises from the uniform partition \(\mathcal{P}=\left\\{x_{0}, x_{1}, \ldots, x_{N}\right\\}\) of \(I,\) we use the left endpoint \(x_{j-1}\) instead of the right endpoint in formula \((5.1 .9),\) then we obtain the left endpoint approximation $$ \sum_{j=1}^{N} f\left(x_{j-1}\right) \cdot \Delta x $$ of the area \(A\) of the region below the graph of \(f\) and above \(I\) (see Figure 10 ). As can be seen from Figure \(10,\) if \(f\) increases on \(I,\) then the right endpoint approximation overestimates \(A,\) and the left endpoint approximation underestimates \(A .\) In each of Exercises \(39-44,\) calculate the average of the left and right endpoint approximations. (For purposes of comparison, the exact value of \(A\) is given. Notice that your answer is more accurate than both the left and right endpoint approximations.) $$ f(x)=\sin (x) \quad I=[0, \pi / 3], N=2, A=1 / 2 $$

Step-by-Step Solution

Answer

The average of the left and right endpoint approximations is \(\frac{\pi(1 + \sqrt{3})}{24}\).

Step 1: Determine the Partition Points

Given the interval \(I = [0, \pi/3]\) and \(N = 2\), we divide the interval into 2 equal subintervals. This means each subinterval has a width \(\Delta x = \frac{\pi/3 - 0}{2} = \pi/6\). The partition points are \(x_0 = 0\), \(x_1 = \pi/6\), and \(x_2 = \pi/3\).

Step 2: Calculate Left Endpoint Approximation

For the left endpoint approximation, we evaluate \(f(x)\) at the left endpoints of the subintervals. Thus, we have:\[ f(x_0) = \sin(0) = 0 \]\[ f(x_1) = \sin(\pi/6) = \frac{1}{2} \]Then, the left endpoint approximation is:\[ \sum_{j=1}^{2} f(x_{j-1}) \cdot \Delta x = f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x = 0 \cdot \frac{\pi}{6} + \frac{1}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12} \]

Step 3: Calculate Right Endpoint Approximation

For the right endpoint approximation, we evaluate \(f(x)\) at the right endpoints of the subintervals. Thus, we have:\[ f(x_1) = \sin(\pi/6) = \frac{1}{2} \]\[ f(x_2) = \sin(\pi/3) = \frac{\sqrt{3}}{2} \]Then, the right endpoint approximation is:\[ \sum_{j=1}^{2} f(x_j) \cdot \Delta x = f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x = \frac{1}{2} \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2} \cdot \frac{\pi}{6} = \frac{\pi}{12} + \frac{\sqrt{3}}{12}\pi \]

Step 4: Calculate Average of Approximations

To find the average of the left and right endpoint approximations, we use the formula:\[ \text{Average} = \frac{\text{Left Endpoint Approximation} + \text{Right Endpoint Approximation}}{2} \]Substituting the previous results gives:\[ \text{Average} = \frac{\frac{\pi}{12} + \left( \frac{\pi}{12} + \frac{\sqrt{3}}{12}\pi \right)}{2} = \frac{\frac{2\pi}{12} + \frac{\sqrt{3}\pi}{12}}{2} = \frac{\pi + \sqrt{3}\pi}{24} = \frac{\pi(1 + \sqrt{3})}{24} \]

Key Concepts

Integration

Integration is a fundamental concept in calculus, allowing us to accumulate quantities over an interval or find areas under curves. In the context of our exercise, we're dealing with definite integration, which computes the area under a curve given specific boundaries. Here, the interval is between 0 and \(\pi/3\).
This is represented as the integral \(\int_{0}^{\pi/3} \sin(x) \, dx\). The goal of definite integration is to find the exact area under the curve of the function \(f(x) = \sin(x)\) over the given interval. This provides a precise value, which we compare against our approximation methods.
Essentially, integration transforms a summation problem into a continuous form. It allows us to analyze the entirety of a range, rather than just discrete points.
To get the exact area \(A\), one would compute the integral of \(\sin(x)\) from 0 to \(\pi/3\), yielding \(A = \frac{1}{2}\) in this instance.

Approximation Methods

Approximation methods offer a way to estimate integrals when exact solutions are complex or impossible to calculate. One popular approximation technique is using Riemann sums, which come in different forms including left endpoint, right endpoint, and midpoint approximations. Each of these methods provides a different approach to estimating the area under a curve by utilizing discrete partitions of the integration interval.
In our exercise, we use left and right endpoint approximations. Here, the interval \([0, \pi/3]\) is divided into two subintervals with equal width, \(\Delta x = \pi/6\). For the left endpoint approximation, we take the function value at the start of each subinterval, while for the right one, we take the function value at the end.
The left endpoint approximation gives us an underestimate of the area when the function is increasing, while the right approximation tends to overestimate under the same condition. The exact area is often more accurately approximated by taking the average of these two estimates.

Trigonometric Functions

Trigonometric functions, like \(\sin(x)\), play a crucial role in this exercise. They result in periodic oscillations or wave-like patterns. The sine function, in particular, starts at 0, rises to 1 at \(\pi/2\), and returns to 0 at \(\pi\).
Within our specific interval \([0, \pi/3]\), the sine function increases from \(\sin(0) = 0\) to \(\sin(\pi/3) = \sqrt{3}/2\). This behavior influences the outcome of our approximation methods, since the function's upward slope leads the left endpoint sum to be less than the actual area, while the right endpoint sum goes beyond it.
Understanding how trigonometric functions like \(\sin(x)\) behave over specific intervals is essential for correctly applying integration and approximation methods. This knowledge allows us to predict the results of using these methods and adjust accordingly to achieve closer estimates of the true area.