Question

exercises 5 - 8 refer to the region r enclosed between the graph of the function y = 2x - x² and the x - axis for 0 ≤ x ≤ 2. 5. (a) sketch the region r. (b) partition 0, 2 into 4 subintervals and show the four rectangles that lram uses to approximate the area of r. compute the lram sum without a calculator. 6. repeat exercise 1(b) for rram and mram.

Explanation:

5. (a)

1. First, find the roots of the function \(y = 2x - x^{2}\).

• Set \(y = 0\), then \(2x - x^{2}=x(2 - x)=0\). The roots are \(x = 0\) and \(x = 2\).
• Rewrite the function as \(y=-(x^{2}-2x)=-(x - 1)^{2}+1\). This is a parabola opening down - ward with vertex at \((1,1)\). To sketch the region \(R\), plot the parabola \(y = 2x - x^{2}\) for \(0\leq x\leq2\) and shade the region between the parabola and the \(x\) - axis.

5. (b)

1. Partition the interval \([0,2]\) into 4 sub - intervals:

• The width of each sub - interval \(\Delta x=\frac{b - a}{n}\), where \(a = 0\), \(b = 2\), and \(n = 4\). So \(\Delta x=\frac{2-0}{4}=0.5\).
• The sub - intervals are \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), and \([1.5,2]\).

1. For Left - hand Rectangular Approximation Method (LRAM):

• The left - hand endpoints of the sub - intervals are \(x_0 = 0\), \(x_1 = 0.5\), \(x_2 = 1\), \(x_3 = 1.5\).
• Calculate the function values at these endpoints:
• \(y_0=2(0)-(0)^{2}=0\).
• \(y_1=2(0.5)-(0.5)^{2}=1 - 0.25 = 0.75\).
• \(y_2=2(1)-(1)^{2}=1\).
• \(y_3=2(1.5)-(1.5)^{2}=3 - 2.25 = 0.75\).
• The LRAM sum \(A_{LRAM}=\sum_{i = 0}^{3}f(x_i)\Delta x\).
• \(A_{LRAM}=\Delta x(y_0 + y_1+y_2 + y_3)\).
• Substitute \(\Delta x = 0.5\), \(y_0 = 0\), \(y_1 = 0.75\), \(y_2 = 1\), \(y_3 = 0.75\) into the formula:
• \(A_{LRAM}=0.5(0 + 0.75+1 + 0.75)=0.5\times2.5 = 1.25\).

6.

1. For Right - hand Rectangular Approximation Method (RRAM):

• The right - hand endpoints of the sub - intervals \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), \([1.5,2]\) are \(x_1 = 0.5\), \(x_2 = 1\), \(x_3 = 1.5\), \(x_4 = 2\).
• Calculate the function values:
• \(y_1=2(0.5)-(0.5)^{2}=0.75\).
• \(y_2=2(1)-(1)^{2}=1\).
• \(y_3=2(1.5)-(1.5)^{2}=0.75\).
• \(y_4=2(2)-(2)^{2}=0\).
• The RRAM sum \(A_{RRAM}=\sum_{i = 1}^{4}f(x_i)\Delta x\).
• \(A_{RRAM}=\Delta x(y_1 + y_2+y_3 + y_4)\).
• Substitute \(\Delta x = 0.5\), \(y_1 = 0.75\), \(y_2 = 1\), \(y_3 = 0.75\), \(y_4 = 0\) into the formula:
• \(A_{RRAM}=0.5(0.75 + 1+0.75 + 0)=0.5\times2.5 = 1.25\).

1. For Mid - point Rectangular Approximation Method (MRAM):

• The mid - points of the sub - intervals \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), \([1.5,2]\) are \(x_1=0.25\), \(x_2 = 0.75\), \(x_3 = 1.25\), \(x_4 = 1.75\).
• Calculate the function values:
• \(y_1=2(0.25)-(0.25)^{2}=0.5 - 0.0625 = 0.4375\).
• \(y_2=2(0.75)-(0.75)^{2}=1.5 - 0.5625 = 0.9375\).
• \(y_3=2(1.25)-(1.25)^{2}=2.5 - 1.5625 = 0.9375\).
• \(y_4=2(1.75)-(1.75)^{2}=3.5 - 3.0625 = 0.4375\).
• The MRAM sum \(A_{MRAM}=\sum_{i = 1}^{4}f(x_i)\Delta x\).
• \(A_{MRAM}=\Delta x(y_1 + y_2+y_3 + y_4)\).
• Substitute \(\Delta x = 0.5\), \(y_1 = 0.4375\), \(y_2 = 0.9375\), \(y_3 = 0.9375\), \(y_4 = 0.4375\) into the formula:
• \(A_{MRAM}=0.5(0.4375 + 0.9375+0.9375 + 0.4375)=0.5\times2.75 = 1.375\).

Answer:

5. (a)

1. First, find the roots of the function \(y = 2x - x^{2}\).

• Set \(y = 0\), then \(2x - x^{2}=x(2 - x)=0\). The roots are \(x = 0\) and \(x = 2\).
• Rewrite the function as \(y=-(x^{2}-2x)=-(x - 1)^{2}+1\). This is a parabola opening down - ward with vertex at \((1,1)\). To sketch the region \(R\), plot the parabola \(y = 2x - x^{2}\) for \(0\leq x\leq2\) and shade the region between the parabola and the \(x\) - axis.

5. (b)

1. Partition the interval \([0,2]\) into 4 sub - intervals:

• The width of each sub - interval \(\Delta x=\frac{b - a}{n}\), where \(a = 0\), \(b = 2\), and \(n = 4\). So \(\Delta x=\frac{2-0}{4}=0.5\).
• The sub - intervals are \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), and \([1.5,2]\).

1. For Left - hand Rectangular Approximation Method (LRAM):

• The left - hand endpoints of the sub - intervals are \(x_0 = 0\), \(x_1 = 0.5\), \(x_2 = 1\), \(x_3 = 1.5\).
• Calculate the function values at these endpoints:
• \(y_0=2(0)-(0)^{2}=0\).
• \(y_1=2(0.5)-(0.5)^{2}=1 - 0.25 = 0.75\).
• \(y_2=2(1)-(1)^{2}=1\).
• \(y_3=2(1.5)-(1.5)^{2}=3 - 2.25 = 0.75\).
• The LRAM sum \(A_{LRAM}=\sum_{i = 0}^{3}f(x_i)\Delta x\).
• \(A_{LRAM}=\Delta x(y_0 + y_1+y_2 + y_3)\).
• Substitute \(\Delta x = 0.5\), \(y_0 = 0\), \(y_1 = 0.75\), \(y_2 = 1\), \(y_3 = 0.75\) into the formula:
• \(A_{LRAM}=0.5(0 + 0.75+1 + 0.75)=0.5\times2.5 = 1.25\).

6.

1. For Right - hand Rectangular Approximation Method (RRAM):

• The right - hand endpoints of the sub - intervals \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), \([1.5,2]\) are \(x_1 = 0.5\), \(x_2 = 1\), \(x_3 = 1.5\), \(x_4 = 2\).
• Calculate the function values:
• \(y_1=2(0.5)-(0.5)^{2}=0.75\).
• \(y_2=2(1)-(1)^{2}=1\).
• \(y_3=2(1.5)-(1.5)^{2}=0.75\).
• \(y_4=2(2)-(2)^{2}=0\).
• The RRAM sum \(A_{RRAM}=\sum_{i = 1}^{4}f(x_i)\Delta x\).
• \(A_{RRAM}=\Delta x(y_1 + y_2+y_3 + y_4)\).
• Substitute \(\Delta x = 0.5\), \(y_1 = 0.75\), \(y_2 = 1\), \(y_3 = 0.75\), \(y_4 = 0\) into the formula:
• \(A_{RRAM}=0.5(0.75 + 1+0.75 + 0)=0.5\times2.5 = 1.25\).

1. For Mid - point Rectangular Approximation Method (MRAM):

• The mid - points of the sub - intervals \([0,0.5]\), \([0.5,1]\), \([1,1.5]\), \([1.5,2]\) are \(x_1=0.25\), \(x_2 = 0.75\), \(x_3 = 1.25\), \(x_4 = 1.75\).
• Calculate the function values:
• \(y_1=2(0.25)-(0.25)^{2}=0.5 - 0.0625 = 0.4375\).
• \(y_2=2(0.75)-(0.75)^{2}=1.5 - 0.5625 = 0.9375\).
• \(y_3=2(1.25)-(1.25)^{2}=2.5 - 1.5625 = 0.9375\).
• \(y_4=2(1.75)-(1.75)^{2}=3.5 - 3.0625 = 0.4375\).
• The MRAM sum \(A_{MRAM}=\sum_{i = 1}^{4}f(x_i)\Delta x\).
• \(A_{MRAM}=\Delta x(y_1 + y_2+y_3 + y_4)\).
• Substitute \(\Delta x = 0.5\), \(y_1 = 0.4375\), \(y_2 = 0.9375\), \(y_3 = 0.9375\), \(y_4 = 0.4375\) into the formula:
• \(A_{MRAM}=0.5(0.4375 + 0.9375+0.9375 + 0.4375)=0.5\times2.75 = 1.375\).