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In Exercises \(1-4,\) use the Midpoint Rule with \(n=4\) to approximate the area of the region. Compare your result with the exact area obtained with a definite integral. $$ f(x)=-2 x+3, \quad[0,1] $$

Short Answer

Expert verified
The solution involves calculating an integral approximation using the Midpoint Rule and making a comparison with the exact integration results.

Step by step solution

01

Define Function and Interval

Here the function \(f(x)=-2x+3\), and the interval \([a, b] = [0, 1]\). The number of subintervals \(n\) is given as 4.
02

Calculate the Width of Subintervals

The width of each subinterval \(\Delta x\) is calculated by the formula \(\Delta x = (b - a) / n\). Substituting values we get \(\Delta x = (1 - 0) / 4 = 0.25\).
03

Apply Midpoint Rule

According to the Midpoint Rule, the approximate integral is given by \(\Delta x \times \sum_{i=1}^{n}f(mid_i)\) where \(mid_i\) is the midpoint for each subinterval. The snippet would look like this: \(\Delta x[f(.125) + f(.375) + f(.625) + f(.875)]\). Calculate the function value at each midpoint and make necessary computations.
04

Calculate Exact Definite Integral

The exact value can be obtained through definite integration over the interval [0, 1]. The integral \(\int_{0}^{1} (-2x + 3) dx = -x^{2} + 3x\Big|_0^1 = 1\).
05

Compare Midpoint Approximation with Exact Area

Having both the approximation using the Midpoint Rule and the exact value through definite integration, we can now make a comparison to reflect the accuracy of the Midpoint Rule.

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Key Concepts

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

Definite Integral
A definite integral is a numerical value representing the area under a curve between two points. In the context of our example, it refers to finding the area under the function \( f(x) = -2x + 3 \) from \( x = 0 \) to \( x = 1 \). The definite integral is written mathematically as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
For our function, the calculation involves finding the antiderivative, which would be \(-x^2 + 3x\), and then evaluating it from 0 to 1. Performing the calculation, we find:
\[ \int_{0}^{1} (-2x + 3) \, dx = [-x^2 + 3x]_0^1 = 1 \]
This tells us that the exact area under the curve from 0 to 1 is 1 square unit.
Approximation Methods
Approximation methods are techniques used to estimate values that cannot be calculated exactly. In calculus, when determining the area under a curve, approaches like the Midpoint Rule can provide a good approximation when an exact integral is complex or undesirable to compute.
Using approximation methods can be incredibly useful when:
  • The function is difficult to integrate analytically.
  • An approximate result is adequate for a given application.
  • We want to compare an approximation against calculated exact values for error analysis.
In our example, we use the Midpoint Rule where the interval \([0, 1]\) is divided into 4 subintervals. For each subinterval, the midpoint is used to approximate the area under the curve.
This method involves these steps:
  • Divide the region into an equal number of subintervals.
  • Use the midpoint of each subinterval to evaluate the function.
  • Multiply the result by the width of the subinterval.
In practice, this process generates a reasonably accurate estimate especially if the interval is well-aligned with the curve of the function.
Numerical Integration
Numerical integration involves approximating the value of a definite integral using techniques that convert the area under a curve into numerical values. Compared to symbolic integration, which finds the exact area using calculus, numerical integration techniques like the Midpoint Rule, Trapezoidal Rule, or Simpson's Rule, simplify this process by breaking it down into more manageable calculations.
Here are some key points about numerical integration:
  • It's essential for dealing with complex functions that lack easy antiderivatives.
  • Provides a way to work with real-world data where functions might be non-analytic.
  • Numerical results can be sufficient for computational purposes and analyses.
In our instance, using the Midpoint Rule to approximate the integral of \( f(x) = -2x + 3 \) gives us a practical numerical result. Although not exact, it allows us to manage and manipulate data that might otherwise be challenging to handle using traditional calculus methods.

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Most popular questions from this chapter

The table gives the marginal benefit and marginal cost of producing \(x\) units of a product for a given company. Plot the points in each column and use the regression feature of a graphing utility to find a linear model for marginal benefit and a quadratic model for marginal cost. Then use integration to find the benefit \(B\) and cost \(C\) equations. Assume \(B(0)=0\) and \(C(0)=425 .\) Finally, find the intervals in which the benefit exceeds the cost of producing \(x\) units, and make a recommendation for how many units the company should produce based on your findings. (Source: Adapted from Taylor, Economics, Fifth Edition) \(\begin{array}{|l|c|c|c|c|}\hline \text { Number of units } & {1} & {2} & {3} & {4} & {5} \\ \hline \text { Marginal benefit } & {330} & {320} & {290} & {270} & {250} \\ \hline \text { Marginal cost } & {150} & {120} & {100} & {110} & {120} \\ \hline\end{array}\) \(\begin{array}{|l|l|l|l|l|}\hline \text { Number of units } & {6} & {7} & {8} & {9} & {10} \\ \hline \text { Marginal benefit } & {230} & {210} & {190} & {170} & {160} \\ \hline \text { Marginal cost } & {140} & {160} & {190} & {250} & {320} \\ \hline\end{array}\)

A company produces a product for which the marginal cost of producing \(x\) units is modeled by \(d C / d x=2 x-12,\) and the fixed costs are dollar 125 . (a) Find the total cost function and the average cost function. (b) Find the total cost of producing 50 units. (c) In part (b), how much of the total cost is fixed? How much is variable? Give examples of fixed costs associated with the manufacturing of a product. Give examples of variable costs.

Evaluate the definite integral. \(\int_{0}^{4}\left(x^{1 / 2}+x^{1 / 4}\right) d x\)

Sketch the region whose area is represented by the definite integral. Then use a geometric formula to evaluate the integral. \(\int_{-3}^{3} \sqrt{9-x^{2}} d x\)

Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all -values in the interval for which the function is equal to its average value. Function \(\quad\) Interval \(f(x)=x \sqrt{4-x^{2}} \quad[0,2]\)

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