Problem 71 - Calculus Single Variable

Question

The partition \\{0,15,25,50,75,90,100\\} of [0,100] given in Exercise 46 precludes an application of Simpson's Rule because it is not uniform. However, because the number of subintervals is even, the idea behind Simpson's Rule can be used. Find the degree 2 polynomial \(p_{1}\) that passes through the three points \((0,0),(15,3),\) and \((25,19) .\) The interpolation command of a computer algebra system will make short work of finding this polynomial. Calculate \(\int_{0}^{25} p_{1}(x) d x .\) Next, find the degree 2 polynomial \(p_{2}\) that passes through the three points \((25,19),(50,25),\) and \((75,42) .\) Calculate \(\int_{25}^{75} p_{2}(x) d x .\) Repeat this procedure with the last three points, \((75,42),(90,75),\) and (100,100) . What approximation of \(\int_{0}^{100} L(x) d x\) results?

Step-by-Step Solution

Answer

The approximation of \(\int_{0}^{100} L(x) dx\) is 265.28.

Step 1: Find the Polynomial through Points (0,0),(15,3),(25,19)

To find the degree 2 polynomial, we assume a polynomial of the form \(p_1(x) = ax^2 + bx + c\). Using the points, we set up the equations: \(p_1(0) = 0\), \(p_1(15) = 3\), and \(p_1(25) = 19\). This gives us three equations:1. \(c = 0\)2. \(a(15)^2 + b(15) + 0 = 3\)3. \(a(25)^2 + b(25) + 0 = 19\).Solving these equations, we find \(a = \frac{2}{75}\), \(b = -\frac{1}{5}\), and \(c = 0\). Thus, \(p_1(x) = \frac{2}{75}x^2 - \frac{1}{5}x\).

Step 2: Calculate \(\int_{0}^{25} p_{1}(x) \, dx\)

Integrate the polynomial \(p_1(x) = \frac{2}{75}x^2 - \frac{1}{5}x\) from 0 to 25:\[ \int_{0}^{25} \left( \frac{2}{75}x^2 - \frac{1}{5}x \right) dx = \left[ \frac{2}{75}\cdot\frac{x^3}{3} - \frac{1}{5}\cdot\frac{x^2}{2} \right]_0^{25}\]Evaluating, this gives:\[ \left( \frac{2}{225}\cdot25^3 - \frac{1}{10}\cdot25^2 \right) - 0 = \frac{6250}{225} - 62.5 = \frac{25}{9} \approx 2.78 \]

Step 3: Find the Polynomial through Points (25,19),(50,25),(75,42)

Assume \(p_2(x) = ax^2 + bx + c\). Using these points:1. \(a(25)^2 + b(25) + c = 19\)2. \(a(50)^2 + b(50) + c = 25\)3. \(a(75)^2 + b(75) + c = 42\).Solving, we get \(a = \frac{11}{375}\), \(b = -\frac{1}{75}\), \(c = 0\). Thus, \(p_2(x) = \frac{11}{375}x^2 - \frac{1}{75}x\).

Step 4: Calculate \(\int_{25}^{75} p_{2}(x) \, dx\)

Integrate \(p_2(x) = \frac{11}{375}x^2 - \frac{1}{75}x\) from 25 to 75:\[ \int_{25}^{75} \left( \frac{11}{375}x^2 - \frac{1}{75}x \right) dx = \left[ \frac{11}{375}\cdot\frac{x^3}{3} - \frac{1}{75}\cdot\frac{x^2}{2} \right]_{25}^{75}\]Evaluating, results in:\[ \left( \frac{11}{1125}\cdot75^3 - \frac{1}{150}\cdot75^2 \right) - \left( \frac{11}{1125}\cdot25^3 - \frac{1}{150}\cdot25^2 \right) = 25 \]

Step 5: Find the Polynomial through Points (75,42),(90,75),(100,100)

Assume \(p_3(x) = ax^2 + bx + c\). Using the points, set up the following:1. \(a(75)^2 + b(75) + c = 42\)2. \(a(90)^2 + b(90) + c = 75\)3. \(a(100)^2 + b(100) + c = 100\).Solving gives \(a = \frac{3}{625}\), \(b = \frac{21}{125}\), \(c = -87\). Thus, \(p_3(x) = \frac{3}{625}x^2 + \frac{21}{125}x - 87\).

Step 6: Calculate \(\int_{75}^{100} p_{3}(x) \, dx\)

Integrate \(p_3(x) = \frac{3}{625}x^2 + \frac{21}{125}x - 87\) from 75 to 100:\[ \int_{75}^{100} \left( \frac{3}{625}x^2 + \frac{21}{125}x - 87 \right) dx = \left[ \frac{3}{625}\cdot\frac{x^3}{3} + \frac{21}{125}\cdot\frac{x^2}{2} - 87x \right]_{75}^{100}\]Evaluating gives:\[ 237.5 \]

Step 7: Sum the Integrals to Approximate \(\int_{0}^{100} L(x) dx\)

Add the results from the three intervals to get the total approximation:\[ \int_{0}^{25} p_1(x) \, dx + \int_{25}^{75} p_2(x) \, dx + \int_{75}^{100} p_3(x) \, dx = 2.78 + 25 + 237.5 = 265.28 \]

Key Concepts

Polynomial Interpolation

Polynomial interpolation helps us find a polynomial that fits a set of points as closely as possible. In simpler terms, imagine you have several dots on a graph, and you want to draw a smooth curve passing exactly through these dots. This process is called interpolation. The polynomial found through interpolation is used to estimate values between the given points.
In the exercise, degree 2 polynomials (quadratic polynomials) are used for interpolation. We use the general form of a quadratic polynomial, which is \(p(x) = ax^2 + bx + c\). By plugging in the x and y values of the points, we can set up a system of equations to solve for the coefficients \(a\), \(b\), and \(c\).

• For each set of three points, a unique quadratic polynomial is determined.
• This polynomial can be used for various calculations, like finding the area under the curve.

Understanding polynomial interpolation is crucial as it is the foundation for many mathematical and real-world applications. It allows us to make educated guesses about data points that lie within the range of known values.

Numerical Integration

Numerical integration techniques are methods used to compute the integral of a function when an analytical solution is challenging or impossible to find. It involves breaking down the area under a curve into simpler shapes (like rectangles or trapezoids) that can be calculated easily.
In this exercise, while not using Simpson's Rule directly due to non-uniform intervals, a similar approach is applied. The exercise finds polynomials that approximate the segments of the curve, and then these polynomials are integrated over specific intervals. This process provides the approximate area, representing the integral of the function.

• Numerical integration allows us to estimate integrals without finding a closed-form solution.
• It is valuable in scenarios where data points are discrete, irregular, or when dealing with complex functions.

Using numerical integration enables us to draw conclusions about the behavior of a function over an interval through approximate calculations.

Integral Approximation

Integral approximation is about estimating the value of an integral when a precise calculation is not feasible. This is crucial in real-world problems where a perfect solution is not only difficult but sometimes unnecessary. The exercise in the original problem involves estimating the integral \( \int_{0}^{100} L(x) \, dx \) by summing up the integrals over different intervals.
By using quadratic polynomials to approximate parts of the function \( L(x) \), the solution effectively breaks down the calculation into manageable chunk that can be easily integrated. Each part of the function is approximated separately, and these results are summed up to provide the final approximation.

• The idea is to combine simple, calculated areas to form a total approximation.
• This method helps to understand and solve complex integration problems with simpler mathematical tools.

Integral approximation is a powerful technique to infer properties and values of functions in a pragmatic and computationally efficient manner.