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Chapter 5: Problem 5

Use the following table to estimate \(\int_{0}^{15} f(x) d x\). $$ \begin{array}{c|c|c|c|c|c|c} \hline x & 0 & 3 & 6 & 9 & 12 & 15 \\ \hline f(x) & 50 & 48 & 44 & 36 & 24 & 8 \\ \hline \end{array} $$

Short Answer

Expert verified
The estimated value of the integral is 543.

Step by step solution

01

Understand the Setup

We need to estimate the integral of the function \( f(x) \) over the interval from \( x = 0 \) to \( x = 15 \). We will use the given table of values for \( f(x) \) to approximate this integral.
02

Choose the Method

We will use the trapezoidal rule to approximate the integral based on the table values. This rule is suited for estimating definite integrals when only discrete data points are available.
03

Apply the Trapezoidal Rule Formula

The trapezoidal rule is given by: \[\int_{a}^{b} f(x) \, dx \approx \frac{1}{2} \sum_{i=1}^{n}(x_i - x_{i-1})(f(x_{i-1}) + f(x_i))\] where \(n\) is the number of subintervals. In our case, the subintervals are \([0, 3]\), \([3, 6]\), \([6, 9]\), \([9, 12]\), \([12, 15]\), with five subintervals.
04

Calculate Each Subinterval Contribution

- For \([0, 3]\): \( \frac{1}{2} \cdot (3 - 0) \cdot (50 + 48) = \frac{1}{2} \cdot 3 \cdot 98 = 147 \)- For \([3, 6]\): \( \frac{1}{2} \cdot (6 - 3) \cdot (48 + 44) = \frac{1}{2} \cdot 3 \cdot 92 = 138 \)- For \([6, 9]\): \( \frac{1}{2} \cdot (9 - 6) \cdot (44 + 36) = \frac{1}{2} \cdot 3 \cdot 80 = 120 \)- For \([9, 12]\): \( \frac{1}{2} \cdot (12 - 9) \cdot (36 + 24) = \frac{1}{2} \cdot 3 \cdot 60 = 90 \)- For \([12, 15]\): \( \frac{1}{2} \cdot (15 - 12) \cdot (24 + 8) = \frac{1}{2} \cdot 3 \cdot 32 = 48 \)
05

Sum Up the Contributions

Add all the contributions together to get the approximate value of the integral: \[ 147 + 138 + 120 + 90 + 48 = 543 \]

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Key Concepts

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

Understanding Numerical Integration
Numerical integration is a powerful mathematical technique that helps us calculate approximate values of definite integrals.
It is especially useful when the function under investigation is complex or we only have function values at discrete data points.
Unlike traditional integration methods that require an explicit function, numerical integration focuses on estimate techniques using available data.
  • It's ideal for situations where a function is challenging to integrate symbolically.
  • Instead of trying to find an antiderivative, approximations build over discrete segments of the domain.
  • This approach is vital in applied sciences, where exact function forms aren't always known.
In our example, we are presented with a set of points, and our goal is to find the definite integral of these points over a defined range using established numerical methods.
Exploring Definite Integral and its Importance
A definite integral represents the area under a curve over a specified range. Simply put, it sums up infinite tiny segments from a function's graph to find the total accumulated value over an interval.
This visualization aids in understanding real-world problems like finding distances, probabilities, and accumulated changes. In our scenario,
  • We aim to calculate the integral from 0 to 15, signifying the total area between the curve defined by our data points and the x-axis.
  • The resulting value gives us insights into quantity or change accumulated across these points.
  • This measure is independent of how scattered or irregular the function may appear.
The task utilizes numerical methods like the trapezoidal rule to derive the definite integral from available discrete data.
Making Use of Discrete Data Points
In numerical integration, discrete data points serve as the primary building blocks for our calculations. These points may come from experimental data, computed algorithms, or measurements.
With discrete data, we have values for a specific set of inputs rather than a continuous function formula.
  • The trapezoidal rule, used in this example, leverages these discrete points to draw many tiny adjacent trapezoids under the curve and sum their areas.
  • Given our table, the data points at specific x-values are crucial in forming these approximations.
  • These data-driven methods are foundational for computations in many scientific fields, where continuous data curves aren't accessible.
By understanding how discrete data points fit into the broader context of numerical analysis, learners can better appreciate how approximations are made without explicit function equations.

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Most popular questions from this chapter

The rate of change of a quantity is given by \(f(t)=t^{2}+1\). Make an underestimate and an overestimate of the total change in the quantity between \(t=0\) and \(t=8\) using (a) \(\quad \Delta t=4\) (b) \(\Delta t=2\) (c) \(\Delta t=1\) What is \(n\) in each case? Graph \(f(t)\) and shade rectangles to represent each of your six answers.

Ice is forming on a pond at a rate given by $$ \frac{d y}{d t}=\frac{\sqrt{t}}{2} \text { inches per hour, } $$ where \(y\) is the thickness of the ice in inches at time \(t\) measured in hours since the ice started forming. (a) Estimate the thickness of the ice after 8 hours. (b) At what rate is the thickness of the ice increasing after 8 hours?

(a) Use a graph of the integrand to make a rough estimate of the integral. Explain your reasoning. (b) Use a computer or calculator to find the value of the definite integral. $$ \int_{0}^{1} 3^{t} d t $$

The value of a mutual fund increases at a rate of \(R=\) \(500 e^{0.04 t}\) dollars per year, with \(t\) in years since \(2010 .\) (a) Using \(t=0,2,4,6,8,10\), make a table of values for \(R\). (b) Use the table to estimate the total change in the value of the mutual fund between 2010 and 2020 .

The rate of change of the world's population, in millions of people per year, is given in the following table. (a) Use this data to estimate the total change in the world's population between 1950 and \(2000 .\) (b) The world population was 2555 million people in 1950 and 6085 million people in \(2000 .\) Calculate the true value of the total change in the population. How does this compare with your estimate in part (a)? \begin{equation} \begin{array}{c|c|c|c|c|c|c} \hline \text { Year } & 1950 & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline \text { Rate of change } & 37 & 41 & 78 & 77 & 86 & 79 \\ \hline \end{array} \end{equation}
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