Problem 1 - University Calculus: Early Transcendentals

Question

Use finite approximations to estimate the area under the graph of the function using a. a lower sum with two rectangles of equal width. b. a lower sum with four rectangles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width. \(f(x)=x^{2}\) between \(x=0\) and \(x=1.\)

Step-by-Step Solution

Answer

Lower sum for two rectangles: 0.125, four rectangles: 0.21875. Upper sum for two rectangles: 0.625, four rectangles: 0.46875.

Step 1: Determine Rectangle Width for Two Rectangles

To find the area using two rectangles, divide the interval from 0 to 1 into two equal parts. The width of each rectangle for two parts is \( \frac{1-0}{2} = 0.5 \).

Step 2: Lower Sum Approximation with Two Rectangles

For the lower sum using two rectangles, choose the left endpoint for each subinterval to ensure heights are based on the function's minimum value in each subinterval. - The first rectangle is based on \( f(0) = 0^2 = 0 \) with width 0.5, giving area = \( 0 \times 0.5 = 0 \).- The second rectangle is based on \( f(0.5) = (0.5)^2 = 0.25 \) with width 0.5, giving area = \( 0.25 \times 0.5 = 0.125 \). Total lower sum = 0 + 0.125 = 0.125.

Step 3: Lower Sum Approximation with Four Rectangles

For lower sum using four rectangles, calculate the width: \( \frac{1-0}{4} = 0.25 \).- The first rectangle is based on \( f(0) = 0^2 = 0 \) with width 0.25, area = \( 0 \times 0.25 = 0 \).- The second rectangle is based on \( f(0.25) = (0.25)^2 = 0.0625 \) with width 0.25, area = \( 0.0625 \times 0.25 = 0.015625 \).- The third rectangle is based on \( f(0.5) = (0.5)^2 = 0.25 \) with width 0.25, area = \( 0.25 \times 0.25 = 0.0625 \).- The fourth rectangle is based on \( f(0.75) = (0.75)^2 = 0.5625 \) with width 0.25, area = \( 0.5625 \times 0.25 = 0.140625 \).Total lower sum = 0 + 0.015625 + 0.0625 + 0.140625 = 0.21875.

Step 4: Upper Sum Approximation with Two Rectangles

For the upper sum using two rectangles, choose the right endpoint for each subinterval to ensure heights are based on the function's maximum value in each subinterval. - The first rectangle is based on \( f(0.5) = (0.5)^2 = 0.25 \) with width 0.5, area = \( 0.25 \times 0.5 = 0.125 \).- The second rectangle is based on \( f(1) = 1^2 = 1 \) with width 0.5, area = \( 1 \times 0.5 = 0.5 \).Total upper sum = 0.125 + 0.5 = 0.625.

Step 5: Upper Sum Approximation with Four Rectangles

For the upper sum using four rectangles, the width remains \( 0.25 \).- The first rectangle is based on \( f(0.25) = (0.25)^2 = 0.0625 \) with width 0.25, area = \( 0.0625 \times 0.25 = 0.015625 \).- The second rectangle is based on \( f(0.5) = (0.5)^2 = 0.25 \) with width 0.25, area = \( 0.25 \times 0.25 = 0.0625 \).- The third rectangle is based on \( f(0.75) = (0.75)^2 = 0.5625 \) with width 0.25, area = \( 0.5625 \times 0.25 = 0.140625 \).- The fourth rectangle is based on \( f(1) = 1^2 = 1 \) with width 0.25, area = \( 1 \times 0.25 = 0.25 \).Total upper sum = 0.015625 + 0.0625 + 0.140625 + 0.25 = 0.46875.

Key Concepts

Finite Approximations

Finite approximations are a useful method to estimate the area under a curve when finding the exact area is complex. The basic idea is to use simple geometric shapes like rectangles to approximate the area. Imagine the curve of the function is like a landscape you want to measure. By covering the landscape with rectangles, you closely estimate the total area beneath the curve.
This method is particularly helpful for functions defined on a specific interval, such as the function \( f(x) = x^2 \) between \( x = 0 \) and \( x = 1 \). The area under this curve can be approximated by dividing the interval into equal segments. Rectangles are then used on these segments. The height of each rectangle is determined by the function value at certain points in the interval.
Finite approximations don't provide an exact area, but they give estimates that become more accurate with more rectangles. Understanding and using this approach is an excellent way to start grasping integration and can be essential in many fields of science and engineering.

Lower Sum

The lower sum is a method for estimating the area under a curve using rectangles. For the function \( f(x) = x^2 \) between \( x = 0 \) and \( x = 1 \), we can find a lower sum by placing rectangles whose heights are determined by the minimum function value in each subinterval. This means that in each segment of the divided interval, the rectangle's top won't reach the curve, thus providing a smaller area estimate.
Using two rectangles of equal width, which in this interval is 0.5, we look at two function values: \( f(0) \) and \( f(0.5) \). These values determine the height of our rectangles:

• The first rectangle is based on \( f(0) = 0 \).
• The second is based on \( f(0.5) = 0.25 \).

The lower sum is thus 0.125.
Increasing the number of rectangles to four narrows each to a width of 0.25, using points \( x = 0, 0.25, 0.5, \) and \( 0.75 \) for heights. The more rectangles used, the closer the lower sum approaches the actual area. The total lower sum with four rectangles is 0.21875.

Upper Sum

In contrast to the lower sum, the upper sum uses maximum function values within subintervals to form each rectangle's height. For the function \( f(x) = x^2 \) from \( x = 0 \) to \( x = 1 \), the upper sum involves evaluating the function at the right endpoint of each subinterval.
With two rectangles, each segment of width 0.5, you use the points \( f(0.5) \) and \( f(1) \), resulting in rectangles that slightly overshoot the curve:

• The first rectangle height is from \( f(0.5) = 0.25 \).
• The second rectangle is \( f(1) = 1 \).

Adding these gives an upper sum of 0.625.
Using four rectangles turns the width to 0.25, dividing the interval more finely. This option uses values \( f(0.25), f(0.5), f(0.75), \) and \( f(1) \) as heights, increasing the estimate's accuracy, and achieves a total of 0.46875. The finer the division, the better the upper sum approximates the real area.

Rectangles of Equal Width

Rectangles of equal width are fundamental when using finite approximations to estimate the area under curves. For simplicity and accuracy, the interval of interest is divided into equal sections, which ensures consistent rectangle widths across the entire interval.
In our example of finding the area under the curve of \( f(x) = x^2 \) from \( x = 0 \) to \( x = 1 \), dividing this interval results in basic sections, where each rectangle has the same width. Two rectangles give a width of 0.5 each, while four rectangles provide a width of 0.25 each.
This method is beneficial because it allows for straightforward comparisons between different numbers of rectangles, like comparing results from two versus four rectangles of width. Consistent widths ensure that as more rectangles are added, the approximations become better, reducing the difference between successive lower and upper sums, ultimately getting closer to the true area under the curve.