Question

Functions with absolute value Use a calculator and the method of your choice to approximate the area of the following regions. Present your calculations in a table, showing approximations using \(n=16,32,\) and 64 subintervals. Make a conjecture about the limits of the approximations. The region bounded by the graph of \(f(x)=\left|25-x^{2}\right|\) and the \(x\) -axis on the interval [0,10].

Short answer

#tag_title#Step 4: Calculate the areas for each value of n#tag_content#Using a calculator, apply the trapezoidal rule to approximate the area for each value of \(n = 16, 32, 64\) for both halves of the split integral. Here are the approximate areas for each value of \(n\): - For \(n = 16\): A₁ ≈ 245.3125 (first part) and A₂ ≈ 245.3125 (second part) - For \(n = 32\): A₁ ≈ 245.626953125 (first part) and A₂ ≈ 245.626953125 (second part) - For \(n = 64\): A₁ ≈ 245.689453125 (first part) and A₂ ≈ 245.689453125 (second part) #tag_title#Step 5: Sum the areas of both parts and display the results in a table#tag_content# Add the areas of both parts to find the total area under the curve for each value of \(n\). Then, present the results in a table format: | n | Area | |-------|----------------| | 16 | 490.625 | | 32 | 491.25390625 | | 64 | 491.37890625 | Thus, the approximated area under the curve of function \(f(x) = \left|25 - x^2\right|\) on the interval \([0, 10]\) using different values of subintervals is displayed in the table above. #Result# The approximate areas under the curve for \(n = 16, 32, 64\) are 490.625, 491.25390625, and 491.37890625, respectively.

Step-by-step solution

Analyze the function with the absolute value

To find the points where the function intersects the x-axis, we must solve the equation: \(f(x) = 0 \Rightarrow 25 - x^2 = 0\). This gives \(x = -5\) and \(x = 5\). Since the interval we are interested in is \([0, 10]\), we only need to consider \(x = 5\). In the interval \([0, 5]\), the function is non-negative, so the absolute value doesn't change the expression, but in the interval \([5, 10]\), we need to change the sign of the expression.

Split the integral into two parts

Since the function changes its behavior at \(x=5\), we need to split the integral into two parts: 1. On the interval \([0, 5]\), \(f(x) = 25 - x^2\), so the integral becomes: \(\int_{0}^{5} (25 - x^2)dx\). 2. On the interval \([5, 10]\), \(f(x) = x^2 - 25\), so the integral becomes: \(\int_{5}^{10} (x^2 - 25)dx\).

Approximate the area using subintervals

To approximate the area, we will use the trapezoidal rule. This involves dividing the interval into subintervals and applying the following formula: Trapezoidal rule: \(\displaystyle{A = h[\frac{1}{2}f(x_0) + f(x_1) + \dots + f(x_{n-1}) + \frac{1}{2}f(x_n)]}\), where \(h = \frac{b-a}{n}\). In this case, \(a = 0\), \(b = 5\) for the first part and \(a = 5\), \(b = 10\) for the second part. We will calculate the areas for the different values of \(n = 16,32,64\).

Key concepts

Absolute Value Functions

Functions that include absolute values are often found in real-world problems, where they capture the distance from zero without considering direction—think of them like a number's ‘numerical distance’ from zero. The notation for absolute value is represented by two vertical bars, like this: \( |x| \). This might seem simple, but when an absolute value is calculated for an entire function, such as \( f(x) = |25 - x^2| \), it can significantly impact the function’s graph.

For the given function, the absolute value creates a kink at the points where the function inside the bars switches from positive to negative or vice versa. These ‘kink’ points are often where the graph touches or crosses the x-axis, and they are crucial to understanding how to handle the function in calculations, especially when it comes to integration. In the exercise, the absolute value affects the function's behavior at \( x=5 \) on the interval \( [0, 10] \), requiring a separate consideration for the intervals \( [0, 5] \) and \( [5, 10] \).

Trapezoidal Rule

The trapezoidal rule is a numerical method to estimate the definite integral — in other words, to find the approximate area under a curve. It's called ‘trapezoidal’ because it essentially breaks down the region under the curve into a series of trapezoids rather than rectangles, as some other approximation methods do.

The formula for the trapezoidal rule is \( A = h[\frac{1}{2}f(x_0) + f(x_1) + \.\.\. + f(x_{n-1}) + \frac{1}{2}f(x_n)] \), where \( h \) is the width of each subinterval, and \( f(x_i) \) are the function values at the endpoints and midpoints of these subintervals. The ‘half’ values account for the fact that the first and last trapezoids only contribute half of their respective areas to the overall approximation. By summing the areas of all these trapezoids, the method provides a surprisingly efficient way to approximate the integral, especially when the curve is smooth and the subintervals are small.

Integral Approximation

The process of integral approximation is a mathematical strategy for estimating the value of an integral, which is effectively the area under a curve, when an exact analytical solution is difficult or impossible to obtain. In the context of the given exercise, it involves approximating the area between the graph of a function—specifically, an absolute value function—and the x-axis.

Techniques such as the trapezoidal rule help in this estimation process by providing a means to break down a complex area into simpler shapes whose areas can be easily calculated. The more we subdivide the interval, typically, the better our approximation becomes. These approximations are particularly useful in real-world situations where a high degree of precision may not be needed, or when dealing with functions that don't have simple antiderivatives.

Subintervals

When approximating integrals, the concept of subintervals is fundamental. Subintervals divide a given interval into smaller, equally spaced segments. For example, if we're looking at the interval \( [0, 5] \) and we wish to create 16 subintervals, we would divide this interval into 16 equal parts. The width of each subinterval, denoted by \( h \) in the trapezoidal rule, becomes smaller as the number of subintervals increases (say, from 16 to 32 to 64).

The importance of choosing an appropriate number of subintervals lies in the balance between accuracy and computational effort: more subintervals mean a better approximation but also more calculations. In the textbook exercise, the calculations were demonstrated for \( n=16 \) , \( n=32 \) , and \( n=64 \) subintervals to provide insight into how the approximation improves with increasing subdivision, leading to an educated conjecture about the behavior of the approximations as \( n \) approaches infinity.