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Sure, let's go through the process to estimate [tex]\(\int_3^9 f(x) \, dx\)[/tex] using different Riemann sums and a midpoint sum. Here's a detailed step-by-step solution.
Let's start by organizing the given data in a more readable format:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline f(x) & -3.6 & -2.2 & -0.7 & 0.3 & 0.9 & 1.5 & 1.7 \\ \hline \end{array} \][/tex]
The interval [tex]\([3, 9]\)[/tex] is divided into three equal subintervals, each with a width of [tex]\(2\)[/tex]:
[tex]\[ \Delta x = \frac{9 - 3}{3} = 2 \][/tex]
### Part (a) - Right Riemann Sum
For the right Riemann sum, we take the right endpoints of each subinterval:
- For [tex]\([3, 5]\)[/tex], the right endpoint is [tex]\(5\)[/tex] ([tex]\(f(5) = -0.7\)[/tex])
- For [tex]\([5, 7]\)[/tex], the right endpoint is [tex]\(7\)[/tex] ([tex]\(f(7) = 0.9\)[/tex])
- For [tex]\([7, 9]\)[/tex], the right endpoint is [tex]\(9\)[/tex] ([tex]\(f(9) = 1.7\)[/tex])
So, the right Riemann sum [tex]\(R_3\)[/tex] is:
[tex]\[ R_3 = \Delta x \left[ f(5) + f(7) + f(9) \right] = 2 \left( -0.7 + 0.9 + 1.7 \right) = 2 \left( 1.9 \right) = 3.8 \][/tex]
Since the function is increasing, the right Riemann sum will be greater than the exact value of the integral.
### Part (b) - Left Riemann Sum
For the left Riemann sum, we take the left endpoints of each subinterval:
- For [tex]\([3, 5]\)[/tex], the left endpoint is [tex]\(3\)[/tex] ([tex]\(f(3) = -3.6\)[/tex])
- For [tex]\([5, 7]\)[/tex], the left endpoint is [tex]\(5\)[/tex] ([tex]\(f(5) = -0.7\)[/tex])
- For [tex]\([7, 9]\)[/tex], the left endpoint is [tex]\(7\)[/tex] ([tex]\(f(7) = 0.9\)[/tex])
So, the left Riemann sum [tex]\(L_3\)[/tex] is:
[tex]\[ L_3 = \Delta x \left[ f(3) + f(5) + f(7) \right] = 2 \left( -3.6 + (-0.7) + 0.9 \right) = 2 \left( -3.4 \right) = -6.8 \][/tex]
Since the function is increasing, the left Riemann sum will be less than the exact value of the integral.
### Part (c) - Midpoint Sum
For the midpoint sum, we typically use the midpoints of the subintervals. The midpoints and the corresponding function values are:
- For [tex]\([3, 5]\)[/tex], the midpoint is [tex]\(4\)[/tex] ([tex]\(f(4) = -2.2\)[/tex])
- For [tex]\([5, 7]\)[/tex], the midpoint is [tex]\(6\)[/tex] ([tex]\(f(6) = 0.3\)[/tex])
- For [tex]\([7, 9]\)[/tex], the midpoint is [tex]\(8\)[/tex] ([tex]\(f(8) = 1.5\)[/tex])
So, the midpoint sum [tex]\(M_3\)[/tex] is:
[tex]\[ M_3 = \Delta x \left[ f(4) + f(6) + f(8) \right] = 2 \left( -2.2 + 0.3 + 1.5 \right) = 2 \left( -0.4 \right) = -0.8 \][/tex]
Summarizing the results:
- Right Riemann sum [tex]\(R_3 = 3.8\)[/tex]
- Left Riemann sum [tex]\(L_3 = -6.8\)[/tex]
- Midpoint sum [tex]\(M_3 = -0.8\)[/tex]
With this setup, you have detailed estimates using different methods for the given integral.
Let's start by organizing the given data in a more readable format:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline f(x) & -3.6 & -2.2 & -0.7 & 0.3 & 0.9 & 1.5 & 1.7 \\ \hline \end{array} \][/tex]
The interval [tex]\([3, 9]\)[/tex] is divided into three equal subintervals, each with a width of [tex]\(2\)[/tex]:
[tex]\[ \Delta x = \frac{9 - 3}{3} = 2 \][/tex]
### Part (a) - Right Riemann Sum
For the right Riemann sum, we take the right endpoints of each subinterval:
- For [tex]\([3, 5]\)[/tex], the right endpoint is [tex]\(5\)[/tex] ([tex]\(f(5) = -0.7\)[/tex])
- For [tex]\([5, 7]\)[/tex], the right endpoint is [tex]\(7\)[/tex] ([tex]\(f(7) = 0.9\)[/tex])
- For [tex]\([7, 9]\)[/tex], the right endpoint is [tex]\(9\)[/tex] ([tex]\(f(9) = 1.7\)[/tex])
So, the right Riemann sum [tex]\(R_3\)[/tex] is:
[tex]\[ R_3 = \Delta x \left[ f(5) + f(7) + f(9) \right] = 2 \left( -0.7 + 0.9 + 1.7 \right) = 2 \left( 1.9 \right) = 3.8 \][/tex]
Since the function is increasing, the right Riemann sum will be greater than the exact value of the integral.
### Part (b) - Left Riemann Sum
For the left Riemann sum, we take the left endpoints of each subinterval:
- For [tex]\([3, 5]\)[/tex], the left endpoint is [tex]\(3\)[/tex] ([tex]\(f(3) = -3.6\)[/tex])
- For [tex]\([5, 7]\)[/tex], the left endpoint is [tex]\(5\)[/tex] ([tex]\(f(5) = -0.7\)[/tex])
- For [tex]\([7, 9]\)[/tex], the left endpoint is [tex]\(7\)[/tex] ([tex]\(f(7) = 0.9\)[/tex])
So, the left Riemann sum [tex]\(L_3\)[/tex] is:
[tex]\[ L_3 = \Delta x \left[ f(3) + f(5) + f(7) \right] = 2 \left( -3.6 + (-0.7) + 0.9 \right) = 2 \left( -3.4 \right) = -6.8 \][/tex]
Since the function is increasing, the left Riemann sum will be less than the exact value of the integral.
### Part (c) - Midpoint Sum
For the midpoint sum, we typically use the midpoints of the subintervals. The midpoints and the corresponding function values are:
- For [tex]\([3, 5]\)[/tex], the midpoint is [tex]\(4\)[/tex] ([tex]\(f(4) = -2.2\)[/tex])
- For [tex]\([5, 7]\)[/tex], the midpoint is [tex]\(6\)[/tex] ([tex]\(f(6) = 0.3\)[/tex])
- For [tex]\([7, 9]\)[/tex], the midpoint is [tex]\(8\)[/tex] ([tex]\(f(8) = 1.5\)[/tex])
So, the midpoint sum [tex]\(M_3\)[/tex] is:
[tex]\[ M_3 = \Delta x \left[ f(4) + f(6) + f(8) \right] = 2 \left( -2.2 + 0.3 + 1.5 \right) = 2 \left( -0.4 \right) = -0.8 \][/tex]
Summarizing the results:
- Right Riemann sum [tex]\(R_3 = 3.8\)[/tex]
- Left Riemann sum [tex]\(L_3 = -6.8\)[/tex]
- Midpoint sum [tex]\(M_3 = -0.8\)[/tex]
With this setup, you have detailed estimates using different methods for the given integral.
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