Question

using rectangles whose height is given by the value of the function at the midpoint of the rectangles base, estimate the area under the graph using first two and then four rectangles. f(x) = x² between x = 2 and x = 3 using two rectangles to estimate, the area under f(x) is approximately 101/16 (type an integer or a simplified fraction.) using four rectangles to estimate, the area under f(x) is approximately (type an integer or a simplified fraction.)

Explanation:

Step1: Determine the width of each rectangle

The interval is from \( x = 2 \) to \( x = 3 \), so the length of the interval is \( 3 - 2 = 1 \). For four rectangles, the width \( \Delta x \) of each rectangle is \( \frac{3 - 2}{4}=\frac{1}{4} \).

Step2: Find the midpoints of each sub - interval

The sub - intervals are:

• First sub - interval: \( [2,2+\frac{1}{4}]=[2,\frac{9}{4}] \), midpoint \( x_1 = 2+\frac{1}{8}=\frac{17}{8} \)
• Second sub - interval: \( [\frac{9}{4},\frac{10}{4}] = [\frac{9}{4},\frac{5}{2}] \), midpoint \( x_2=\frac{9}{4}+\frac{1}{8}=\frac{18 + 1}{8}=\frac{19}{8} \)
• Third sub - interval: \( [\frac{5}{2},\frac{11}{4}] \), midpoint \( x_3=\frac{5}{2}+\frac{1}{8}=\frac{20 + 1}{8}=\frac{21}{8} \)
• Fourth sub - interval: \( [\frac{11}{4},3] \), midpoint \( x_4=\frac{11}{4}+\frac{1}{8}=\frac{22+1}{8}=\frac{23}{8} \)

Step3: Calculate the height of each rectangle (function value at midpoint)

The function is \( f(x)=x^{2} \).

• \( f(x_1)=(\frac{17}{8})^{2}=\frac{289}{64} \)
• \( f(x_2)=(\frac{19}{8})^{2}=\frac{361}{64} \)
• \( f(x_3)=(\frac{21}{8})^{2}=\frac{441}{64} \)
• \( f(x_4)=(\frac{23}{8})^{2}=\frac{529}{64} \)

Step4: Calculate the area of each rectangle and sum them up

The area of each rectangle is \( A_i = f(x_i)\Delta x \), and the total area \( A=\sum_{i = 1}^{4}f(x_i)\Delta x \)
Since \( \Delta x=\frac{1}{4} \), we have:

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Answer:

\(\frac{405}{64}\)