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Chapter 4: Problem 10

The Midpoint rule for approximating \(\int_{-1}^{1} f(x) d x\) gives the value 12, the Composite Midpoint rule with \(n=2\) gives 5, and Composite Simpson's rule gives 6 . Use the fact that \(f(-1)=f(1)\) and \(f(-0.5)=f(0.5)-1\) to determine \(f(-1), f(-0.5), f(0), f(0.5)\), and \(f(1)\)

Short Answer

Expert verified
The function values at the respective points are: \(f(-1) = -2\), \(f(-0.5) = -0.5\), \(f(0) = 6\), \(f(0.5) = 0.5\), and \(f(1) = -2\).

Step by step solution

01

Midpoint rule

The Midpoint approximation \(M\) of an integral from \(a\) to \(b\) is given by \((b - a) f((a+b) / 2)\). Since \(a = -1\) and \(b = 1\) and \(M = 12\), it follows that \(2f(0) = 12\). Therefore, \(f(0) = 6\).
02

Composite Midpoint rule

The Composite Midpoint rule with \(n = 2\) intervals is given by \((b - a)/(2n) \[ f(a) + 2f(a+h) + f(b)\]\), where \(h = (b - a)/n\). In this case, \(a = -1\), \(b = 1\), \(n = 2\), and the Composite Midpoint approximation equals 5. Substituting these values and \(f(0) = 6\), we get \(1/2 \[ f(-1) + 2f(0) + f(1) \] = 5\). Since \(f(-1) = f(1)\), we can reduce this to \(f(-1) + 12 = 10\), so \(f(-1) = f(1) = -2\).
03

Use of given condition

We have also the condition \(f(-0.5)=f(0.5)-1\). As \(f(0.5)\) and \(f(-0.5)\) will have equal absolute values but opposite signs, we can conclude \(f(-0.5) = -f(0.5)\). Given \(f(-0.5)=f(0.5)-1\), this can be rewritten as \(-f(0.5) = f(0.5) - 1\), leading to \(2f(0.5) = 1\), thus \(f(0.5) = 0.5\) and \(f(-0.5) = -0.5\).
04

Recap

Given the conditions of integral approximations and the given function conditions, it is found that \(f(-1) = f(1) = -2\), \(f(-0.5) = -0.5\), \(f(0) = 6\), and \(f(0.5) = 0.5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Midpoint Rule
The Midpoint Rule is a simple yet effective method for approximating definite integrals. It is particularly useful when you have polynomial or smooth continuous functions. This rule utilizes the midpoint of the subinterval as the point of evaluation for the function, making it more accurate than simply using an endpoint. The general formula is:
\[M = (b - a) imes f\left(\frac{a+b}{2}\right)\]With \(a = -1\) and \(b = 1\), the formula simplifies to 2\(f(0)\). In this exercise, we know that the Midpoint Rule approximation equals 12, leading to the equation 2\(f(0) = 12\), solving which gives \(f(0) = 6\). The Midpoint Rule is straightforward but becomes less accurate with smaller intervals, thus necessitating its composite version for better precision on wider intervals.
Composite Midpoint Rule
The Composite Midpoint Rule is utilized to improve the accuracy of the Midpoint Rule by dividing the interval into multiple subintervals and applying the Midpoint Rule to each. The number of subintervals is determined by \(n\). The formula for the Composite Midpoint Rule with \(n\) intervals is:
\[\frac{b-a}{2n} \, \left[f(a) + 2\sum_{i=1}^{n-1} f\left(a + ih\right) + f(b)\right]\]where \(h = \frac{b-a}{n}\). In this exercise, with \(n = 2\), substituting in the known values and using the condition \(f(-1) = f(1)\), we simplify to \(f(-1) + 12 = 10\), resulting in \(f(-1) = f(1) = -2\). This method provides a more accurate estimate as it considers more points to gauge the function's behavior over the interval.
Simpson's Rule
Simpson's Rule is another method for approximating the integral of a function and is especially noted for its accuracy. This rule uses a weighted average of function values at both endpoints and the midpoint, effectively approximating the function with quadratic polynomials. This makes it quite powerful for functions that are approximately quadratic in nature over small intervals. The Simpson's Rule formula is:
\[S = \frac{b-a}{6} \, \left[f(a) + 4f\left(\frac{a+b}{2}\right) + f(b)\right]\]In the exercise, the composite Simpson's Rule approximation amounts to 6. Despite its added complexity over the Midpoint and Composite Midpoint Rules, it often provides better accuracy for smooth functions. Using the conditions such as \(f(-0.5) = f(0.5) - 1\) and the values obtained from earlier rules, such as \(f(0) = 6\), Simpson's Rule can help verify these function values' accuracy in practice.

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Most popular questions from this chapter

To simulate the thermal characteristics of disk brakes (see the following figure), D. A. Secrist and R. W. Hornbeck [SH] needed to approximate numerically the "area averaged lining temperature," \(T\), of the brake pad from the equation $$ T=\frac{\int_{r_{e}}^{r_{0}} T(r) r \theta_{p} d r}{\int_{r_{e}}^{r} r \theta_{p} d r} $$ where \(r_{e}\) represents the radius at which the pad-disk contact begins, \(n\) represents the outside radius of the pad-disk contact, \(\theta_{p}\) represents the angle subtended by the sector brake pads, and \(T(r)\) is the temperature at each point of the pad, obtained numerically from analyzing the heat equation (see Section 12.2). Suppose \(r_{e}=0.308 \mathrm{ft}, r_{0}=0.478 \mathrm{ft}, \theta_{p}=0.7051\) radians, and the temperatures given in the following table have been calculated at the various points on the disk. Approximate \(T\). $$ \begin{array}{lccccc} \hline r(\mathrm{ft}) & T(r)\left({ }^{\circ} \mathrm{F}\right) & r(\mathrm{ft}) & T(r)\left({ }^{\circ} \mathrm{F}\right) & r(\mathrm{ft}) & T(r)\left({ }^{\circ} \mathrm{F}\right) \\ \hline 0.308 & 640 & 0.376 & 1034 & 0.444 & 1204 \\ 0.325 & 794 & 0.393 & 1064 & 0.461 & 1222 \\ 0.342 & 885 & 0.410 & 1114 & 0.478 & 1239 \\ 0.359 & 943 & 0.427 & 1152 & & \\ \hline \end{array} $$

Use the following data and the knowledge that the first five derivatives of \(f\) are bounded on \([1,5]\) by \(2,3,6,12\) and 23 , respectively, to approximate \(f^{\prime}(3)\) as accurately as possible. Find a bound for the error. $$ \begin{array}{l|l|l|l|l|l} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 2.4142 & 2.6734 & 2.8974 & 3.0976 & 3.2804 \end{array} $$

Let \(h=(b-a) / 3, x_{0}=a, x_{1}=a+h\), and \(x_{2}=b\). Find the degree of precision of the quadrature formula $$ \int_{a}^{b} f(x) d x=\frac{9}{4} h f\left(x_{1}\right)+\frac{3}{4} h f\left(x_{2}\right) $$.

Approximate \(\int_{0}^{2} x^{2} e^{-x^{2}} d x\) using \(h=0.25\). Use a. Composite Trapezoidal rule. b. Composite Simpson's rule. c. Composite Midpoint rule.

Use Romberg integration to compute the following approximations to $$ \int_{0}^{48} \sqrt{1+(\cos x)^{2}} d x $$ [Note: The results in this exercise are most interesting if you are using a device with between sevenand nine-digit arithmetic.] a. Determine \(R_{1,1}, R_{2,1}, R_{3,1}, R_{4,1}\), and \(R_{5,1}\), and use these approximations to predict the value of the integral. b. Determine \(R_{2,2}, R_{3,3}, R_{4,4}\), and \(R_{55}\), and modify your prediction. c. Determine \(R_{6,1}, R_{6,2}, R_{6,3}, R_{6,4}, R_{6,5}\), and \(R_{6,6}\), and modify your prediction. d. Determine \(R_{7,7}, R_{8,8}, R_{9,9}\), and \(R_{10,10}\), and make a final prediction. e. Explain why this integral causes difficulty with Romberg integration and how it can be reformulated to more easily determine an accurate approximation.
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