Problem 14 - Calculus Single Variable

Question

Calculate the lower and upper Riemann sums for the given function \(f\), interval \(I\), and uniform partition of order 2 $$ f(x)=x^{2}-2 x+2 \quad I=[0,3] $$

Step-by-Step Solution

Answer

Lower sum: 0.75; Upper sum: 11.25.

Step 1: Determine the Partition Points

We are given the interval \( I=[0,3] \) and we need a uniform partition of order 2. This means the interval is divided into 2 equal subintervals. Calculate the width of each subinterval as \[ \Delta x = \frac{b-a}{n} = \frac{3-0}{2} = 1.5. \]Thus, the partition points are \( x_0=0 \), \( x_1=1.5 \), and \( x_2=3 \).

Step 2: Calculate Function Values at Partition Points

We need \( f(x) \) at the partition points. Since \( f(x) = x^2 - 2x + 2 \), calculate:- \( f(0) = 0^2 - 2 \cdot 0 + 2 = 2 \)- \( f(1.5) = (1.5)^2 - 2 \cdot 1.5 + 2 = 0.25 \)- \( f(3) = 3^2 - 2 \cdot 3 + 2 = 5 \)

Step 3: Compute the Lower Riemann Sum

For the lower Riemann sum, choose the minimum function value in each subinterval.- In the interval \([0, 1.5]\), minimum of \( f(0) = 2 \) and \( f(1.5) = 0.25 \) is \( 0.25 \).- In the interval \([1.5, 3]\), minimum of \( f(1.5) = 0.25 \) and \( f(3) = 5 \) is \( 0.25 \).The lower Riemann sum is:\[ L = 0.25 \times 1.5 + 0.25 \times 1.5 = 0.75. \]

Step 4: Compute the Upper Riemann Sum

For the upper Riemann sum, choose the maximum function value in each subinterval.- In the interval \([0, 1.5]\), maximum of \( f(0) = 2 \) and \( f(1.5) = 0.25 \) is \( 2 \).- In the interval \([1.5, 3]\), maximum of \( f(1.5) = 0.25 \) and \( f(3) = 5 \) is \( 5 \).The upper Riemann sum is:\[ U = 2 \times 1.5 + 5 \times 1.5 = 11.25. \]

Key Concepts

Lower Riemann Sum

To understand the Lower Riemann Sum, imagine you are stacking pieces from a puzzle over a curve trying to measure its area. Here, you try to be conservative, choosing the smallest function values over each partition interval. This approach ensures that your calculated 'area' is always less than or equal to the actual area under the curve.
Here’s how it works:

• Divide the interval of interest, in our case \([0,3]\), into smaller subintervals.
• For each subinterval, locate the point that provides the smallest value of the function \(f(x)\).
• Multiply this minimum value by the width of the subinterval.
• Sum these products across all intervals.

For our function \( f(x) = x^2 - 2x + 2 \), and with partition points \([0, 1.5, 3]\), the lower sum selects the minimum value of \(0.25\) within both subintervals. Consequently, the lower Riemann sum is \(0.75\).
This approach ensures the computed area is not overestimating the space beneath the curve.

Upper Riemann Sum

Contrarily, the Upper Riemann Sum approximates the area under a curve by overestimating it. Think of this method as taking a generous step, always choosing the highest values for each subinterval. This results in a sum possibly exceeding the true curve area.
Here’s how it's undertaken:

• Segment the interval \([0,3]\) into smaller divisions.
• In each subinterval, select the maximum value of \( f(x) \).
• Multiply this maximum value by the interval's width.
• Aggregate these products for all subintervals.

For the function \( f(x) = x^2 - 2x + 2 \), and the partition \([0, 1.5, 3]\), the method chooses the maximum values \(2\) and \(5\) for the respective subintervals. The resulting upper sum is \(11.25\), which will be naturally higher than the actual area, providing a safety buffer for underestimation errors elsewhere.

Partition Points

Partition points slice the interval into manageable sections. This concept is crucial, as it lays the groundwork for evaluating the behavior of the function over smaller sections of the domain.
Consider these points like stakes marking sections along a path:

• Start by defining an interval, here \( I = [0,3] \).
• Determine how many subintervals are required. In this case, a uniform partition is required which divides the interval into equal parts, say of order 2.
• Compute the width of each subinterval as \( \Delta x = \frac{b-a}{n} \), where \( n \) is the number of subintervals. For \([0,3]\) with 2 subintervals, our partition points are \([0, 1.5, 3]\).

Partition points set the stage for calculations, ensuring you have well-defined regions where function values can be assessed efficiently.

Piecewise Approximation

Piecewise approximation is like crafting a tiled mosaic to closely mimic a curve's continuous flow with simpler shapes. Each piece represents a segment where calculations become simpler and more manageable.
The strategy involves:

• Taking the defined subintervals, each acting as a piece of the overall approximation scheme.
• In each subinterval, assume the function is either constant or linear depending on the function values selected (minimum or maximum for Riemann sums).
• Use the selected min/max values to draft simple rectangular approximations over those sections.

For \( f(x) = x^2 - 2x + 2 \) in the interval \([0,3]\), each subinterval \([0,1.5]\) and \([1.5,3]\) can be visually imagined as a piece that, when combined, attempts to replicate the overall shape of \( f(x) \). Piecewise approximation helps in breaking down complex tasks, providing a clearer picture through simple, manageable calculations.