let ( f(x) ) be a function with the following table of values:\n\n \begin{array}{c|ccccccc} x = & 1 & 2 & 3…

let ( f(x) ) be a function with the following table of values:\n\n \begin{array}{c|ccccccc} x = & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ f(x) = & 3 & -8 & -5 & -5 & 6 & -2 & -7 end{array} \n\nevaluate the riemann sum of ( f(x) ) on the interval ( 1,7 ) using three equal sub - intervals and left endpoints:\n\n\n\n. (your answer should be a number.)\n\n\n\nquestion 5 (1 point)\n\nconsider a function ( f ) with the following table of values:\n\n \begin{array}{|c|c|c|c|c|c|c|c|} hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ hline f(x) & 4 & -2 & -3 & 0 & 1 & -1 & 2 \\ hline end{array} \n\napproximate ( int_{2}^{6}f(x)dx ) using a riemann sum with two rectangles of equal width and midpoints.

let ( f(x) ) be a function with the following table of values:\n\n \begin{array}{c|ccccccc} x = & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ f(x) = & 3 & -8 & -5 & -5 & 6 & -2 & -7 end{array} \n\nevaluate the riemann sum of ( f(x) ) on the interval ( 1,7 ) using three equal sub - intervals and left endpoints:\n\n\n\n. (your answer should be a number.)\n\n\n\nquestion 5 (1 point)\n\nconsider a function ( f ) with the following table of values:\n\n \begin{array}{|c|c|c|c|c|c|c|c|} hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ hline f(x) & 4 & -2 & -3 & 0 & 1 & -1 & 2 \\ hline end{array} \n\napproximate ( int_{2}^{6}f(x)dx ) using a riemann sum with two rectangles of equal width and midpoints.

Answer

Explanation:

Step1: Calculate the width of sub - intervals

For the first problem: The interval is ([1,7]), (n = 3). The width (\Delta x=\frac{7 - 1}{3}=2). For the second problem: The interval is ([2,6]), (n = 2). The width (\Delta x=\frac{6 - 2}{2}=2).

Step2: Find the left - endpoints (first problem) or mid - points (second problem)

First problem (left - endpoints): The sub - intervals are ([1,3]), ([3,5]), ([5,7]). The left - endpoints are (x_1 = 1), (x_2=3), (x_3 = 5). (f(x_1)=3), (f(x_2)=- 5), (f(x_3)=6). The Riemann sum (S=\sum_{i = 1}^{3}f(x_i)\Delta x=(3-5 + 6)\times2=(4)\times2 = 8). Second problem (mid - points): The sub - intervals are ([2,4]), ([4,6]). The mid - points are (x_1=\frac{2 + 4}{2}=3), (x_2=\frac{4+6}{2}=5). (f(x_1)=0), (f(x_2)=-1). The Riemann sum (S=\sum_{i = 1}^{2}f(x_i)\Delta x=(0-1)\times2=-2).

Answer:

For the first problem: (8) For the second problem: (-2)