Question

1. use the graph of ( y = g(x) ) to estimate ( int_{2}^{10}g(x)dx ) using a left, right, and midpoint riemann sum with ( n = 4 ).

Explanation:

Step1: Calculate the width of each sub - interval

The interval is \([a = 2,b = 10]\) and \(n = 4\). The width of each sub - interval \(\Delta x=\frac{b - a}{n}=\frac{10 - 2}{4}=2\).
The sub - intervals are \([2,4]\), \([4,6]\), \([6,8]\), \([8,10]\)

Step2: Estimate using the left - Riemann sum

For the left - Riemann sum, we use the left - endpoints of each sub - interval: \(x_1 = 2\), \(x_2 = 4\), \(x_3 = 6\), \(x_4 = 8\)
From the graph: \(g(2)\approx - 40\), \(g(4)=0\), \(g(6)=20\), \(g(8)=55\)
\(L_4=\sum_{i = 1}^{4}g(x_{i - 1})\Delta x=\Delta x[g(2)+g(4)+g(6)+g(8)]\)
\(L_4 = 2(-40 + 0+20 + 55)=2\times35 = 70\)

Step3: Estimate using the right - Riemann sum

For the right - Riemann sum, we use the right - endpoints of each sub - interval: \(x_1 = 4\), \(x_2 = 6\), \(x_3 = 8\), \(x_4 = 10\)
From the graph: \(g(4)=0\), \(g(6)=20\), \(g(8)=55\), \(g(10)=20\)
\(R_4=\sum_{i = 1}^{4}g(x_{i})\Delta x=\Delta x[g(4)+g(6)+g(8)+g(10)]\)
\(R_4=2(0 + 20+55 + 20)=2\times95=190\)

Step4: Estimate using the midpoint - Riemann sum

The midpoints of the sub - intervals are \(x_1=\frac{2 + 4}{2}=3\), \(x_2=\frac{4+6}{2}=5\), \(x_3=\frac{6 + 8}{2}=7\), \(x_4=\frac{8+10}{2}=9\)
From the graph: \(g(3)\approx - 20\), \(g(5)=10\), \(g(7)=30\), \(g(9)=55\)
\(M_4=\sum_{i = 1}^{4}g(x_{i})\Delta x=\Delta x[g(3)+g(5)+g(7)+g(9)]\)
\(M_4=2(-20 + 10+30 + 55)=2\times75 = 150\)

Answer:

Left - Riemann sum: \(70\), Right - Riemann sum: \(190\), Midpoint - Riemann sum: \(150\)