Question

In Exercises \(43-48,\) use the limit process to find the area of the region between the graph of the function and the \(x\) -axis over the given interval. Sketch the region. $$ y=16-x^{2}, \quad[1,3] $$

Short answer

The area of the region between the graph of the function \(y=16-x^{2}\) and the \(x\) - axis over the given interval [1,3] is the limit of the given expression in Step 3. The sketching demonstrates this area visually.

Step-by-step solution

Equate the function to zero

This is to locate the x-intercepts of the function. We have \(16 - x^{2} = 0\), from which we get \(x = +\sqrt{16}\) or \(x= - \sqrt{16}\), that is \(x=4\) or \(x=-4\)

Choose the interval

In this step, it is understood that the given interval is \([1, 3]\). This will help in setting up the limits in the calculation of the area

Use the limit process (Riemann Sums)

This step involves summing up the areas of the rectangles formed under the curve of the function in the limit as the width of the rectangles goes to zero. Given function is \(y = 16 - x^{2}\), we choose \(n\) rectangles of equal width \(\Delta x = (b - a) / n = (3 - 1) / n = 2 / n\) within the interval of \([1, 3]\). The area of each rectangle is \(A_i = y_i . \Delta x = (16 - x_i^{2}) . \Delta x\), and \(x_i = a + i.\Delta x = 1 + i. (2/n)\). Use the sum formula for squares, \(1^2 + 2^2 + ... + n^2 = n(n + 1)(2n + 1)/6\), to compute \(\sum_{i=1}^{n}x_i^{2}\) . Hence, we can write the area under the curve as:\[ A = \lim_{n\to\infty}\left[\sum_{i=1}^{n}(16 - (1 + i. (2/n))^{2}) . (2/n)\right] = \lim_{n\to\infty}\left(32 - (1 + 2n + (n(n + 1)(2n + 1)/6n^{2}))(2/n)\right)\]Solving this limit will give the exact area under the curve on the interval \([1, 3]\)

Sketch the function and the area

The function \(y = 16 -x^{2}\) is a downward facing parabola with x-intercepts at \(x = 4\) and \(x = -4\) and maximum point at origin (0,16). The interval \([1, 3]\) on the x-axis is falling on the right hand side of the y-axis. The area under the curve to the x-axis from \(x = 1\) to \(x = 3\) can be shaded to clearly illustrate the area calculation

Key concepts

Riemann Sums

One of the foundational concepts in calculus is Riemann sums, a method for approximating the total area under a curve on a graph, across a certain interval. This technique is essential when it comes to calculating the area when no geometric formula is available.

Riemann sums work by dividing the area into narrow rectangles (the narrower, the better the approximation) and then summing the area of these rectangles. Each rectangle has a width, often denoted as \( \Delta x \), and a height that is determined by the function's value at a specific point within the width of the rectangle.

The choice of this specific point can vary, leading to different types of Riemann sums, such as left, right, and mid-point Riemann sums. The sum of the areas of these rectangles provides an estimate for the total area under the curve that improves as the width of the rectangles decreases.

As the number of rectangles increases indefinitely, and \( \Delta x \) approaches zero, the Riemann sum converges to the exact area. This process is expressed using the limit notation and forms a connection to the concept of integration.

Area under Curve

Determining the area under a curve is a significant application of calculus that has many real-world implications, from calculating the distance traveled over time based on a velocity-time graph to finding the work done against a force over a distance.

The area under a function's curve between two points essentially gives a visual representation of the accumulated sum of values of the function. To translate this into a precise measurement, one must consider the entire set of values the function takes on within the given interval. While geometric shapes like rectangles, triangles, and circles have straightforward area formulas, the area under a more complex curve requires calculus for exact determination.

In mathematics, when students sketch the region between the x-axis and the graph of a function over a given interval, they are visualizing the concept of the area under the curve, which becomes much more tangible and easier to grasp when accompanied by such a visual aid.

Definite Integral Calculus

Definite integral calculus is essentially the formalization of the concept of finding the area under a curve over an interval. This branch of mathematics provides the precise tools necessary to calculate the accumulation of quantities, which can be represented geometrically as the area under a curve on a graph.

The definite integral is symbolized by the integral sign with upper and lower limits representing the interval over which the area is to be found. The use of the definite integral takes the limit of the Riemann sums as the width of the rectangles approaches zero, leading to an exact answer.

The beauty of the definite integral lies in its ability to calculate exact areas irrespective of the shape of the curve or the complexity of the function involved. It is a leap from approximation to precision, where it accounts for all the infinitesimally small pieces of area under the curve. As a result, students can move from the conceptual understanding of Riemann sums to the precise calculations provided by definite integrals.