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Find the Midpoint and Trapezoid Rule approximations to \(\int_{0}^{1} e^{-x} d x\) using \(n=50\) subintervals. Compute the relative error of each approximation.

Short Answer

Expert verified
Also, compute the relative error for each approximation. Answer: The Midpoint Rule approximation is about 0.632275 with a relative error of approximately 0.000245, and the Trapezoid Rule approximation is about 0.632085 with a relative error of approximately 0.000056.

Step by step solution

01

Compute the exact value of the integral

First, let's find the exact value of the definite integral \(\int_{0}^{1} e^{-x} dx\). To do this, we will find the antiderivative of \(e^{-x}\) and then evaluate it at the limits 1 and 0: \(F(x) = \int e^{-x} dx = -e^{-x} + C\) Now, we will compute the definite integral: \(\int_{0}^{1} e^{-x} dx = F(1) - F(0) = -e^{-1} + e^0 = 1 - e^{-1} \approx 0.632120\)
02

Compute the Midpoint Rule approximation

Now, let's find the Midpoint Rule approximation using 50 subintervals: The width of each subinterval is \(h = \frac{1 - 0}{50} = 0.02\). The midpoints are \(x_i = 0.01, 0.03, 0.05, ..., 0.99\). The Midpoint Rule approximation is as follows: \(M_{50} = h \sum_{i=1}^{50} e^{-x_i} \approx 0.02 \sum_{i=1}^{50} e^{-x_i}\) Calculate the sum and multiply by the width to get the Midpoint Rule approximation: \(M_{50} \approx 0.632275\)
03

Compute the Trapezoid Rule approximation

Now, let's find the Trapezoid Rule approximation using 50 subintervals: The Trapezoid Rule approximation is as follows: \(T_{50} = \frac{h}{2} \left[e^{-x_0} + 2e^{-x_1} + 2e^{-x_2} + ... + 2e^{-x_{49}} + e^{-x_{50}}\right]\) Since \(x_0 = 0\) and \(x_{50} = 1\), we can plug these values into the Trapezoid Rule approximation formula: \(T_{50} \approx \frac{0.02}{2} \left[e^0 + 2\sum_{i=1}^{49} e^{-x_i} + e^{-1}\right]\) Calculate the sum and plug into the formula to get the Trapezoid Rule approximation: \(T_{50} \approx 0.632085\)
04

Compute the relative error for each approximation

Finally, let's compute the relative error for each approximation: Relative Error for Midpoint Rule = \(\frac{|M_{50} - \int_{0}^{1} e^{-x} dx|}{\int_{0}^{1} e^{-x} dx}\) Relative Error for Midpoint Rule \(\approx \frac{|0.632275 - 0.632120|}{0.632120} \approx 0.000245\) Relative Error for Trapezoid Rule = \(\frac{|T_{50} - \int_{0}^{1} e^{-x} dx|}{\int_{0}^{1} e^{-x} dx}\) Relative Error for Trapezoid Rule \(\approx \frac{|0.632085 - 0.632120|}{0.632120} \approx 0.000056\) The Midpoint Rule approximation is about 0.632275 with a relative error of approximately \(0.000245\) and the Trapezoid Rule approximation is about 0.632085 with a relative error of approximately \(0.000056\).

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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 numerical method for approximating definite integrals. It works by dividing the area under a curve into a series of rectangles, whose height is determined by the function value at the midpoint of each subinterval. This method is particularly useful when you want an estimate using equally spaced intervals.

For our example, we divided the interval from 0 to 1 into 50 subintervals, each with a width of \(h = 0.02\). The midpoints are at \(x_i = 0.01, 0.03, 0.05, \ldots, 0.99\). By calculating the function value \(e^{-x_i}\) at each midpoint and summing these values multiplied by the subinterval width \(h\), we get the Midpoint Rule approximation. This produces a result close to the actual integral, with our example yielding approximately 0.632275. This approach gives a good estimate, especially when more subintervals are used.
Trapezoid Rule
The Trapezoid Rule is another method for approximating definite integrals, which employs trapezoids instead of rectangles. This technique assumes the function is linear between consecutive points and calculates the integral as a sum of areas of trapezoids formed under the curve.

In our exercise, we also used 50 subintervals. The function is evaluated at the endpoints, forming points \(x_0 = 0\) and \(x_{50} = 1\), and the approximate area under the curve is calculated using the formula:
  • \(T_{50} = \frac{0.02}{2} \left[e^0 + 2\sum_{i=1}^{49} e^{-x_i} + e^{-1}\right]\)
The Trapezoid Rule thus provides an approximation of 0.632085 for the integral. This method can often be more accurate than the Midpoint Rule, especially with a finer subinterval split.
definite integral
A definite integral computes the accumulation of values of a function over a specified interval. Essentially, it finds the net area under a curve, which can represent various physical quantities like distance, area, or volume in real-life scenarios.

In our problem, we find the definite integral of \(e^{-x}\) from 0 to 1. The exact solution can be found by determining the antiderivative, which for \(e^{-x}\) is \(-e^{-x}\). Evaluating this from 1 to 0 gives us the exact integral value, \(1 - e^{-1} \approx 0.632120\). Understanding definite integrals is central to many applications in mathematics and science, providing exact measurements where necessary.
relative error
Relative error helps gauge the accuracy of an approximation by comparing it to the exact value. It gives an idea of how close the numerical approximation is to the true value of the integral.

Calculating the relative error involves finding the difference between the approximate value and the exact integral, divided by the exact value:
  • Relative Error for Midpoint Rule: \(\frac{|M_{50} - \int_{0}^{1} e^{-x} dx|}{\int_{0}^{1} e^{-x} dx}\)
  • Relative Error for Trapezoid Rule: \(\frac{|T_{50} - \int_{0}^{1} e^{-x} dx|}{\int_{0}^{1} e^{-x} dx}\)
For the Midpoint Rule, the relative error calculated is approximately 0.000245, and for the Trapezoid Rule, it is approximately 0.000056. These numbers indicate the Trapezoid Rule approximation is slightly closer to the true value, showing how relative error can guide improvements in numerical methods.

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Most popular questions from this chapter

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)

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Use numerical methods or a calculator to approximate the following integrals as closely as possible. $$\int_{0}^{\infty} \ln \left(\frac{e^{x}+1}{e^{x}-1}\right) d x=\frac{\pi^{2}}{4}$$

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty).\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

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