/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Let \(\Delta x=(b-a) / 2 .\) Use... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 61

Let \(\Delta x=(b-a) / 2 .\) Use the Mean Value Theorem to find points \(s_{1}\) in \([a,(a+b) / 2]\) and \(s_{2}\) in \([(a+b) / 2, b]\) such that the Riemann sum $$ \mathcal{R}\left(f,\left\\{s_{1}, s_{2}\right\\}\right)=f\left(s_{1}\right) \cdot \Delta x+f\left(s_{2}\right) \cdot \Delta x $$ is equal to the Riemann integral of \(f\) over \([a, b]\). $$ f(x)=1 / x \quad [1,2] $$

Short Answer

Expert verified
The points are \(s_1 = \frac{1}{2\ln(1.5)}\) in \([1, 1.5]\) and \(s_2 = \frac{1}{2(\ln(2) - \ln(1.5))}\) in \([1.5, 2]\).

Step by step solution

01

Understanding the Problem

We need to find points using the Mean Value Theorem for integration, such that the Riemann sum is equal to the definite integral of a function over an interval. Here, the function is given as \(f(x) = \frac{1}{x}\) over the interval \([1, 2]\). The interval is split into two parts, \([1, 1.5]\) and \([1.5, 2]\), and we need to use the Mean Value Theorem to find the points \(s_1\) and \(s_2\) where the function equals the average value over these subintervals.
02

Calculating \(\Delta x\)

The term \(\Delta x\) is defined as \((b-a)/2\). Here, \(a = 1\) and \(b = 2\), so we have \(\Delta x = \frac{(2 - 1)}{2} = 0.5\).
03

Setting Up the Mean Value Theorem for Integration

The Mean Value Theorem for integration states that for a continuous function \(f(x)\), there exists a point \(c\) in \([a, b]\) where \(f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx\). We apply this to the subintervals \([1, 1.5]\) and \([1.5, 2]\) to find \(s_1\) and \(s_2\).
04

Calculating the Integral over Subinterval \([1, 1.5]\)

Calculate \(\int_1^{1.5} \frac{1}{x} \, dx\). The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). So, we have \(\int_1^{1.5} \frac{1}{x} \, dx = [\ln x]_1^{1.5} = \ln(1.5) - \ln(1) = \ln(1.5)\).
05

Calculating \(s_1\) Using the Mean Value Theorem

For the subinterval \([1, 1.5]\), set \(f(s_1) \cdot 0.5 = \ln(1.5)\). Therefore, \(\frac{1}{s_1} \cdot 0.5 = \ln(1.5)\), which implies \(s_1 = \frac{1}{2\ln(1.5)}\).
06

Calculating the Integral over Subinterval \([1.5, 2]\)

Calculate \(\int_{1.5}^2 \frac{1}{x} \, dx\). Using the antiderivative \(\ln|x|\), we find \(\int_{1.5}^2 \frac{1}{x} \, dx = [\ln x]_{1.5}^2 = \ln(2) - \ln(1.5)\).
07

Calculating \(s_2\) Using the Mean Value Theorem

For the subinterval \([1.5, 2]\), set \(f(s_2) \cdot 0.5 = \ln(2) - \ln(1.5)\). Therefore, \(\frac{1}{s_2} \cdot 0.5 = \ln(2) - \ln(1.5)\), which implies \(s_2 = \frac{1}{2(\ln(2) - \ln(1.5))}\).
08

Checking the Riemann Sum Equals the Integral

Compute \(f(s_1)\cdot \Delta x + f(s_2)\cdot \Delta x\) using the values of \(s_1\) and \(s_2\) found. Verify if this equals \(\int_1^2 \frac{1}{x} \, dx = \ln(2)\). Since both parts have been calculated such that their evaluated products and sums match the integral over the full interval, the solution is verified.

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Key Concepts

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

Riemann Sum
The Riemann Sum is a fundamental concept in calculus used to approximate the area under a curve or, in other words, the integral of a function. It works by dividing the region under the curve into smaller sections called subintervals.
By using these subintervals, we can create rectangles that approximate the area.
Each rectangle鈥檚 height is determined by the function value at a specific point within the interval.
  • The Riemann Sum is calculated as the sum of the area of these rectangles: \[ \mathcal{R}(f, \{s_1, s_2, ..., s_n\}) = f(s_1) \cdot \Delta x + f(s_2) \cdot \Delta x + ... + f(s_n) \cdot \Delta x \]
  • In our specific example, we were tasked with calculating the sum for two rectangles, which means we needed two function values, \(f(s_1)\) and \(f(s_2)\).
This sum provides us an approximation, but when the number of rectangles becomes infinitely large, the Riemann Sum approaches the exact integral value of the function over the interval.
Definite Integral
The Definite Integral is a crucial idea bridging the Riemann Sum and the exact area under a curve. It gives us the precise accumulation of quantities, such as area, when taking infinitely many infinitely thin slices.
For a continuous function \(f(x)\) over an interval \([a, b]\), the definite integral is represented as:
  • \( \int_a^b f(x) \, dx \)
  • The integral computes the exact area below the curve and above the x-axis over \([a, b]\).
  • In the problem context, we are finding the definite integral of the function \(f(x) = \frac{1}{x}\) over \([1, 2]\).
By using the Mean Value Theorem for Integration, we ensured that the Riemann Sum equals this definite integral, allowing us to find precise function values \(s_1\) and \(s_2\) where this match occurs.
Antiderivative
The Antiderivative, also known as an indefinite integral, is a function \(F(x)\) whose derivative is the given function \(f(x)\). This concept is integral to solving definite integrals as it provides a way to express cumulative values in calculus.
For example, for \(f(x) = \frac{1}{x}\), the antiderivative is:
  • \( F(x) = \ln|x| + C \)
  • \(C\) is any constant, which typically cancels out in definite integral calculations.
In the exercise, knowing that \( \int \frac{1}{x} \, dx = \ln|x| \) allowed us to calculate the specific definite integral values required over the given subintervals \([1, 1.5]\) and \([1.5, 2]\).
These integral results were then employed to locate \(s_1\) and \(s_2\).
Subintervals
Subintervals are smaller divisions of a given interval, used in the approximation of integrals through Riemann Sums and definite integrals. In problems involving integration, the main interval is divided into several parts, each called a subinterval, which simplifies the calculation method by focusing on smaller, manageable sections.
In the context of our exercise:
  • The interval \([1, 2]\) was divided into two subintervals: \([1, 1.5]\) and \([1.5, 2]\).
  • Subdividing aids in utilizing the Mean Value Theorem for Integration more effectively, allowing the calculation of specific points and cumulative areas.
By computing integrals and Riemann sums over these subintervals individually, we can get closer to matching the exact integral value of the whole interval. This approach is instrumental in achieving precise solutions in problems like the one provided.

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Most popular questions from this chapter

A function \(f\) and an interval \([a, b]\) are specified. Calculate the Simpson's Rule approximations of \(\int_{a}^{b} f(x) d x\) with \(N=10\) and \(N=20 .\) If the first five decimal places do not agree, increment \(N\) by \(10 .\) Continue until the first five decimal places of two consecutive approximations are the same. State your answer rounded to four decimal places. $$ f(x)=\sin (\pi \cos (x)) \quad[0, \pi / 3] $$

During a 6 -minute span, at intervals of 1 minute, the speedometer of a car reads $$ \begin{array}{|l|c|c|c|c|c|c|c|}\hline \text { Time (minutes) } & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline \text { Speed (km/hr) } & 90 & 80 & 75 & 80 & 80 & 70 & 60 \\\\\hline\end{array}$$ 3 Use Simpson's Rule to estimate the distance traveled by the car during that 6 -minute period.

In a particular regional climate, the temperature varies between \(-22^{\circ} \mathrm{C}\) and \(40^{\circ} \mathrm{C}\), averaging \(\mu=11^{\circ} \mathrm{C}\). The number of days \(F(T)\) in the year on which the temperature remains below \(T\) degrees centigrade is given (approximately) by $$F(T)=\int_{-22}^{T} f(x) d x \quad(-22 \leq T \leq 40)$$ where $$f(x)=12.72 \exp \left(-\frac{(x-11)^{2}}{266.4}\right) $$ Notice that \(F(T)\) is the sort of area integral that we studied in Section 5.4 . a. Use Simpson's Rule with \(N=20\) to approximate \(F(40) .\) What should the exact value of \(F(40)\) be? b. Heat alerts are issued when the daily high temperature is \(36^{\circ} \mathrm{C}\) or more. On about how many days a year are heat alerts issued? c. Suppose that global warming raises the average temperature by \(1^{\circ} \mathrm{C}\), shifting the graph of \(f\) by 1 unit to the right. The new model may be obtained by simply replacing \(\mu\) with 12 and using [-21,41] as the domain (see Figure 13). What is the percentage increase in heat alerts that will result from this \(1^{\circ} \mathrm{C}\) shift in temperature?

A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=\cos (x) \quad I=[\pi / 4,5 \pi / 4], N=4 $$

The graphs of \(y=f(x)\) and \(y=g(x)\) intersect in more than two points. Find the total area of the regions that are bounded above and below by the graphs of \(f\) and \(g\). $$ f(x)=2 \sin (x) \quad g(x)=6(\pi-x) /(5 \pi) $$
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