Problem 56 - Essential Calculus

Question

In Exercises \(53-56,\) use the Midpoint Rule Area \(\approx \sum_{i=1}^{n} f\left(\frac{x_{i}+x_{i-1}}{2}\right) \Delta x\) with \(n=4\) to approximate the area of the region bounded by the graph of the function and the \(x\) -axis over the given interval. $$ f(x)=\sin x, \quad\left[0, \frac{\pi}{2}\right] $$

Step-by-Step Solution

Answer

Our approximation for the area of the region bound by the graph of \( \sin x \) and the \( x \) -axis over the interval [0, \(\frac{\pi}{2}\)] using the Midpoint Rule with \( n = 4 \) is \(\pi/8 * \left(\sin(\frac{\pi}{8}) + \sin(\frac{3\pi}{8}) + \sin(\frac{5\pi}{8}) + \sin(\frac{7\pi}{8})\right)\).

Step 1: Finding the Width \(\Delta x\)

The given interval is [0, \(\pi/2\)]. Given that \(n = 4\), this interval will be split into four equals parts. So the width of each rectangle or partition on the interval will be \(\Delta x = (\frac{\pi}{2} - 0) / 4 = \frac{\pi}{8}\).

Step 2: Applying the Midpoint Rule to approximate the Area

Calculate \(f\left(\frac{x_{i}+x_{i-1}}{2}\right)\) for each rectangle and then sum up all those values. A simpler method here would be to notice that the midpoints of the intervals are x = \(\frac{\pi}{8}\), x = \(\frac{3\pi}{8}\), x = \(\frac{5\pi}{8}\), and x = \(\frac{7\pi}{8}\), and these are the points that will be entered into the function \(f(x)\). Calculate \(f(x)\) at each midpoint and sum up these values.

Step 3: Performing the Calculations

Calculate the values at each midpoint, so \(f(\frac{\pi}{8}) + f(\frac{3\pi}{8}) + f(\frac{5\pi}{8}) + f(\frac{7\pi}{8}) = \sin(\frac{\pi}{8}) + \sin(\frac{3\pi}{8}) + \sin(\frac{5\pi}{8}) + \sin(\frac{7\pi}{8}). \) Now we know area \(\approx \sum_{i=1}^{n} f\left(\frac{x_{i}+x_{i-1}}{2}\right) \Delta x = \Delta x * \left(\sin(\frac{\pi}{8}) + \sin(\frac{3\pi}{8}) + \sin(\frac{5\pi}{8}) + \sin(\frac{7\pi}{8})\right)\)

Key Concepts

Approximate Area Under Curve

Determining the area beneath a curve on a graph is a staple of calculus, specifically integral calculus. However, finding the exact area can often be challenging or even impossible using elementary methods. Hence, we turn to approximation methods. One such method is the Midpoint Rule, which is a type of Riemann Sum. This approach approximates the region under a curve by dividing it into a finite number of rectangles, using the midpoint of each subinterval as the point at which the function's height is evaluated.

The real value lies in its simplicity and ease of use for functions where evaluating the exact integral is difficult. By increasing the number of rectangles (the variable 'n' in the midpoint formula), we refine our approximation. In our exercise, with the function being \( f(x)=\sin x \) over the interval \( [0, \frac{\pi}{2}] \) and 4 subsections, the Midpoint Rule provides a reasonably accurate estimate for the area, appealing to students new to calculus who are learning to transition from summation to integration.

Riemann Sum

A Riemann Sum is one of the most fundamental concepts in integral calculus and provides a strategy to approximate the total area under a curve—often called the integral—by summing up areas of shapes, generally rectangles, that cover the region. The sum is named after the German mathematician Bernhard Riemann, a key figure in the formalization of integral calculus.

The idea is to partition the interval into smaller subintervals, measure the height of the function at certain points within these intervals—such as left endpoints, right endpoints, or midpoints—and then multiply each height by the width of the subinterval. By summing all these products, we acquire an approximation of the area under the curve. The accuracy of the estimate improves as the width of the subintervals gets smaller, hence the common approach to take the limit as the number of subintervals goes to infinity to find the exact integral.

Definite Integrals

The definite integral is a concept that gives us the exact area under a curve over a specific interval, rather than an approximation. This integral is a foundational piece of calculus and is denoted by an integral symbol with upper and lower limits that specify the interval of interest.

The formal calculation involves taking the limit of a Riemann Sum as the number of subintervals approaches infinity, which yields the precise area. In essence, the definite integral gives us the accumulated quantity—from area to other physical applications—where functions represent rates of change or distributions. While the Midpoint Rule and other Riemann Sums are only approximations, the definite integral, often found by using anti-derivatives, serves as the true value these approximations aspire to reach.

Trigonometric Functions

In our example, the function involved is \( f(x) = \sin x \)—a fundamental trigonometric function expressing the ratio of the length of the side of a right-angled triangle opposite to an angle to the length of the triangle’s hypotenuse. Trigonometric functions are periodic and oscillate between fixed values. The function \( \sin x \) in particular varies between -1 and 1 and is commonly used in both pure and applied mathematics, including physics and engineering.

In the context of integral calculus, trigonometric functions often have to be integrated to find areas, volumes, and other physical quantities. The identity properties and periodic nature of these functions can sometimes simplify integration, particularly if the interval is aligned with the function's period. However, in intervals where the function needs to be estimated, methods like the Midpoint Rule become valuable tools for approximation, as demonstrated with our \( f(x) = \sin x \) example.