Problem 29 - Calculus Early Transcendentals

Question

Complete the following steps for the given function, interval, and value of \(n\). a. Sketch the graph of the function on the given interval. b. Calculate \(\Delta x\) and the grid points \(x_{0}, x_{1}, \ldots, x_{n}\) c. Illustrate the midpoint Riemann sum by sketching the appropriate rectangles. d. Calculate the midpoint Riemann sum. $$f(x)=2 x+1 \text { on }[0,4] ; n=4$$.

Step-by-Step Solution

Answer

Answer: The midpoint Riemann sum is 20.

Step 1: Sketch the Graph of the Function

First, let's sketch the graph of the function \(f(x) = 2x + 1\) on the interval [0, 4]. Since it is a linear function, we only need two points to draw the graph. Let's use the endpoints of the interval. For \(x=0\), we have \(f(0) = 2(0) + 1 = 1\). For \(x=4\), we have \(f(4) = 2(4) + 1 = 9\). Now, draw a straight line connecting these two points (1,0) and (4,9) on a graph.

Step 2: Calculate Delta x and Grid Points

Now, let's find \(\Delta x\) and grid points \(x_{0}, x_{1}, \ldots, x_{n}\). Since we have \(n=4\) partitions, we can find \(\Delta x\) using the following formula: \(\Delta x = \frac{b-a}{n} = \frac{4-0}{4} = 1\). The grid points can be found by starting from \(x_0 = a = 0\) and moving to the right by \(\Delta x = 1\), giving us the grid points \(x_0 = 0\), \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\), and \(x_4 = 4\).

Step 3: Illustrate the Midpoint Riemann Sum

To illustrate the midpoint Riemann sum, we need to find the midpoint of each partition. The midpoints are \(x^*_i =\frac{x_{i-1}+x_i}{2}\); \(x^*_1 = \frac{1}{2}\), \(x^*_2 = \frac{3}{2}\), \(x^*_3 = \frac{5}{2}\), and \(x^*_4 = \frac{7}{2}\). Then, for each midpoint \(x^*_i\), draw a rectangle with a height of \(f(x^*_i)\) and a width of \(\Delta x = 1\).

Step 4: Calculate the Midpoint Riemann Sum

Finally, let's calculate the midpoint Riemann sum. The midpoint Riemann sum is given by the formula: $$M_n = \Delta x \sum_{i=1}^{n} f(x^*_i)$$ So, we first need to find \(f(x^*_i)\): - \(f(x^*_1) = f(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 2\) - \(f(x^*_2) = f(\frac{3}{2}) = 2(\frac{3}{2}) + 1 = 4\) - \(f(x^*_3) = f(\frac{5}{2}) = 2(\frac{5}{2}) + 1 = 6\) - \(f(x^*_4) = f(\frac{7}{2}) = 2(\frac{7}{2}) + 1 = 8\) Now, plug these values into the formula to find the midpoint Riemann sum: $$M_n = 1(2+4+6+8) = 20$$ Thus, the midpoint Riemann sum for the function \(f(x) = 2x + 1\) on the interval \([0,4]\) with \(n=4\) partitions is 20.

Key Concepts

Linear Functions

A linear function is a simple yet powerful tool in mathematics represented as a straight line when graphed. It takes the form \( f(x) = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. The slope \( m \) indicates how steep the line is, and the y-intercept \( c \) shows where the line crosses the y-axis.

Linear functions map every x-value to a particular y-value via a constant rate of change. For example, in the function \( f(x) = 2x + 1 \), the slope is 2, meaning for every increase of 1 unit in \( x \), the value of \( f(x) \) goes up by 2 units. The y-intercept in this case is 1, indicating the line crosses the y-axis at (0,1).

A key characteristic of linear functions is their predictability and simplicity:

• The graph is always a straight line.
• They have consistent slope across the line.
• Easy to calculate values for any \( x \).

With these traits, linear functions serve as a fundamental concept in algebra and calculus, forming the foundation for understanding more complex functions.

Riemann Sums

Riemann sums are a method for approximating the area under a curve. This technique breaks down the problem of finding an area related to a function into manageable rectangles. By adding up the areas of these rectangles, we can estimate the total area under the curve.

Among the different types of Riemann sums (left, right, midpoint), the midpoint Riemann sum uses the midpoint of each subinterval within a partition to calculate the height of rectangles. For the function \( f(x) = 2x + 1 \), on the interval \([0, 4]\) with \( n = 4 \) subdivisions, the midpoints \( x^*_i \) such as \( x^*_1 = 0.5 \), \( x^*_2 = 1.5 \), etc., are used for these calculations.

Using midpoint Riemann sums offers several benefits:

• Provides a more balanced estimate compared to left or right sums.
• Reduces errors in irregular functions by considering average heights.

This approach is essential for transitioning from approximate to precise calculations as seen in integrals.

Numerical Integration

Numerical integration is a critical concept in calculus and involves calculating an area when an analytic integral is difficult or impossible. Techniques like the Riemann sum method help achieve this by approximating the integral or total area under a curve.

Numerical integration is incredibly versatile: it can be applied to continuous functions or discreetly sampled data. The midpoint rule, featured in our exercise, is one of the numerical methods that estimates the integral by computing the sum of midpoint rectangle areas. For the linear function \( f(x) = 2x + 1 \), the midpoint Riemann sum offers a good estimate of the integral over the interval \([0, 4]\).

Understanding numerical integration is crucial:

• It helps deal with complex functions or real-life data calculations.
• Allows for solving integrals when traditional methods can't be applied.
• Forms the basis for other advanced numerical methods in sciences and engineering.

As an approximation tool, it bridges the gap between theoretical calculus and practical application.