/*! 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 21 Estimate each definite integral ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 21

Estimate each definite integral "by hand," using Simpson's Rule with \(n=4\). Round all calculations to three decimal places. Exercises \(19-26\) correspond to Exercises \(1-8\), in which the same integrals were estimated using trapezoids. If you did the corresponding exercise, compare your Simpson's Rule answer with your trapezoidal answer. $$ \int_{2}^{4} \frac{1}{x} d x $$

Short Answer

Expert verified
The integral \( \int_{2}^{4} \frac{1}{x} \, dx \) is approximately 0.693 using Simpson's Rule.

Step by step solution

01

Understand Simpson's Rule

Simpson's Rule is a method for numerical integration that approximates the integral of a function. It works by approximating the region under the curve using parabolas instead of straight lines (as in the Trapezoidal Rule). The formula is: \[ S_n = \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4) \right] \] where \( n \) is even, \( \Delta x = \frac{b-a}{n} \), and \( x_i = a + i\Delta x \).
02

Calculate \(\Delta x\) and \(x_i\)

The interval is from 2 to 4, so set \(a=2\) and \(b=4\). The number of subintervals, \(n\), is 4. Therefore, \( \Delta x = \frac{4-2}{4} = 0.5 \). Calculate the \(x_i\) values: \(x_0 = 2\), \(x_1 = 2.5\), \(x_2 = 3\), \(x_3 = 3.5\), and \(x_4 = 4\).
03

Evaluate the function at \(x_i\) values

For each \(x_i\), evaluate the function \( f(x) = \frac{1}{x} \): - \(f(x_0) = f(2) = \frac{1}{2} = 0.500\) - \(f(x_1) = f(2.5) = \frac{1}{2.5} = 0.400\) - \(f(x_2) = f(3) = \frac{1}{3} = 0.333\) - \(f(x_3) = f(3.5) = \frac{1}{3.5} = 0.286\) - \(f(x_4) = f(4) = \frac{1}{4} = 0.250\)
04

Apply Simpson's Rule formula

Substitute the values into Simpson's Rule formula: \[ S_4 = \frac{0.5}{3} \left[ 0.500 + 4(0.400) + 2(0.333) + 4(0.286) + 0.250 \right] \] = \[ \frac{0.5}{3} \left[ 0.500 + 1.600 + 0.666 + 1.144 + 0.250 \right] \] = \[ \frac{0.5}{3} \times 4.16 \] = 0.693. Therefore, the estimated integral using Simpson's Rule is approximately 0.693.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Numerical integration
Numerical integration is a fundamental technique used when it is difficult or impossible to calculate the integral of a function analytically. This approach is particularly useful for integrals that do not have elementary antiderivatives or when working with data points instead of functions.

Two common methods for numerical integration are Simpson鈥檚 Rule and the Trapezoidal Rule. These methods provide approximate solutions by dividing the integration interval into smaller subintervals and calculating the area under the curve using simple geometrical shapes.

Simpson鈥檚 Rule uses parabolic segments to approximate the curve, which usually gives a more accurate result compared to the trapezoidal approach, especially if the function is smooth and continuous. Understanding these methods allows us to estimate the area under a curve within a given range accurately.
Definite integral
A definite integral, denoted by the integral symbol with upper and lower limits, such as \( \int_{a}^{b} f(x) \, dx \), represents the signed area under the curve of a function \( f(x) \) from \( x = a \) to \( x = b \).

In practical terms, a definite integral calculates the accumulation of quantities, like area, over a specific interval.

The concepts of definite integrals are widely used in physics and engineering for calculating things like work, energy, and charge. Knowing how to approximate definite integrals is invaluable when the integrals are too complex to solve with traditional calculus methods. Numerical integration techniques like Simpson's Rule and the Trapezoidal Rule come into play in these scenarios, providing an efficient way to approximate the true value of the integral.
Trapezoidal Rule
The Trapezoidal Rule is a straightforward numerical integration technique. It approximates the area under a curve by dividing the interval into smaller subintervals and using trapezoids to estimate the sections between the curve and the x-axis.

To apply the Trapezoidal Rule, you divide the interval between two points, \( a \) and \( b \), into \( n \) equal segments. Then, you calculate the average of the function's values at the endpoints of these segments. The formula for the Trapezoidal Rule is: \[ T_n = \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n) \right] \] where \( \Delta x = \frac{b-a}{n} \).

While the Trapezoidal Rule may be less precise than Simpson's Rule for functions that curve significantly, it is a powerful and simple method when dealing with linear or slightly curved functions. With enough subdivisions, it can approximate the integral quite accurately, making it a useful tool in many practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \text { Use integration by parts to find each integral. } $$ $$ \int(x+2) e^{x} d x \quad[\text { Hint } \text { : Take } u=x+2 .] $$

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. Contamination is leaking from an underground waste-disposal tank at the rate of \(4 \ln t\) thousand gallons per month, where \(t\) is the number of months since the leak began. Find the total leakage from the end of month 1 to the end of month \(4 .\)

Suppose that you meet 30 new people each year, but each year you forget \(20 \%\) of all of the people that you know. If \(y(t)\) is the total number of people who you remember after \(t\) years, then \(y\) satisfies the differential equation \(y^{\prime}=30-0.2 y .\) (Do you see why?) Solve this differential equation subject to the condition \(y(0)=0 \quad\) (you knew no one at birth).

Hospital patients are often given glucose (blood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this "drip" supplies glucose at the rate of \(25 \mathrm{mg}\) per minute, and each minute \(10 \%\) of the accumulated glucose is consumed by the body. Then the amount \(y(t)\) of glucose (in excess of the normal level) in the body after \(t\) minutes satisfies $$y^{\prime}=25-0.1 y \quad \text { (Do you see why?) }$$ \(y(0)=0 \quad\) (zero excess glucose at \(t=0)\) Solve this differential equation and initial condition.

We omit the constant of integration when we integrate \(d v\) to get \(v\). Including the constant \(C\) in this step simply replaces \(v\) by \(v+C\), giving the formula $$ \int u d v=u(v+C)-\int(v+C) d u $$ Multiplying out the parentheses and expanding the last integral into two gives $$ \int u d v=u v+C u-\int v d u-C \int d u $$ Show that the second and fourth terms on the right cancel, giving the "old" integration by parts formula \(\int u d v=u v-\int v d u\). This shows that including the constant in the \(d v\) to \(v\) step gives the same formula. One constant of integration at the end is enough. Repeated Integration by Parts Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). $$ \begin{array}{c} \int \underbrace{x^{2} e^{x} d x}_{u d v}=\underbrace{x^{2} e^{x}}_{u v}-\int \underbrace{e^{x} 2 x d x}_{w \atop d u}=x^{2} e^{x}-2 \int x e^{x} d x \\ {\left[\begin{array}{cc} u=x^{2} & d v & =e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x} \end{array}\right]} \end{array} $$ The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) :
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.