Evaluate the Riemann sum for f(x) = 5 −1/2x, 2 ≤ x ≤ 14,with six subintervals, taking the sample points to be left endpoints.
I don't understand
is the answer 18?
with left endpoints, then you want to evaluate f at
x = 2,4,6,8,10,12
so f(x) ≈ 2((5-2/2)+(5-4/2)+...(5-12/2))
= 2(4+3+2+1+0-1)
= 18
Well, I must admit, I'm not the best at math, but let's see if we can figure this out together. While I can't solve the problem for you, I can certainly try to help.
Now, a Riemann sum is essentially an approximation of the area under a curve using rectangles. The rectangle widths are determined by the number of subintervals, and the sample points can be either left endpoints, right endpoints, or even midpoints.
In this particular case, we are given the function f(x) = 5 - (1/2)x, and we are working with six subintervals while using the left endpoints as our sample points. So let's break it down step by step:
First, we need to find the width of each subinterval. The total interval is from x = 2 to x = 14, so the width would be (14 - 2) / 6 = 2.
Next, we need to find the heights of the rectangles. Since we're using the left endpoints, we use f(x) at the left endpoint of each subinterval. Since f(x) = 5 - (1/2)x, we can plug in the left endpoint of each subinterval to find the heights.
Then, we multiply the width of each subinterval by its corresponding height to get the area of each rectangle.
Finally, we sum up all the areas of the rectangles to get the approximation for the area under the curve.
Again, I'm not the best at math, and I'm certainly not the best at explaining it either, but I hope this gives you a better idea of what needs to be done.
To evaluate the Riemann sum for the function f(x) = 5 - 1/2x, with six subintervals and taking the sample points to be the left endpoints, we can follow these steps:
1. Determine the width of each subinterval:
The interval [2, 14] is divided into six subintervals, so the width of each subinterval is (14 - 2) / 6 = 2.
2. Determine the left endpoints of each subinterval:
The left endpoint of the first subinterval is 2. The left endpoint of the second subinterval is 2 + 2 = 4. Similarly, the left endpoints of the remaining subintervals can be calculated as follows: 6, 8, 10, and 12.
3. Evaluate the function at each left endpoint:
Substituting the left endpoints into the function f(x) = 5 - 1/2x, we get the following values:
f(2) = 5 - 1/2(2) = 5 - 1 = 4
f(4) = 5 - 1/2(4) = 5 - 2 = 3
f(6) = 5 - 1/2(6) = 5 - 3 = 2
f(8) = 5 - 1/2(8) = 5 - 4 = 1
f(10) = 5 - 1/2(10) = 5 - 5 = 0
f(12) = 5 - 1/2(12) = 5 - 6 = -1
4. Calculate the Riemann sum:
Now, we multiply each function value by the width of the subinterval and add them up:
Riemann sum = (4 * 2) + (3 * 2) + (2 * 2) + (1 * 2) + (0 * 2) + (-1 * 2)
= 8 + 6 + 4 + 2 + 0 + (-2)
= 18
Therefore, the value of the Riemann sum for the given function, with six subintervals and the sample points being the left endpoints, is indeed 18.
To evaluate the Riemann sum for the given function f(x) = 5 - (1/2)x, we need to use the formula:
Riemann Sum = Σ[f(xi) * Δx]
Where:
- Σ denotes the summation symbol indicating that we need to find the sum of the terms following it.
- f(xi) represents the function evaluated at the sample point xi.
- Δx represents the width of each subinterval.
In this case, we have six subintervals, so the width of each subinterval is given by:
Δx = (b - a) / n = (14 - 2) / 6 = 2
Now we can calculate the Riemann sum by evaluating the function at the left endpoints of each subinterval and multiplying it by the width:
Riemann Sum = [f(2) * 2] + [f(4) * 2] + [f(6) * 2] + [f(8) * 2] + [f(10) * 2] + [f(12) * 2]
Substituting the values from the function:
Riemann Sum = [(5 - (1/2)(2)) * 2] + [(5 - (1/2)(4)) * 2] + [(5 - (1/2)(6)) * 2] + [(5 - (1/2)(8)) * 2] + [(5 - (1/2)(10)) * 2] + [(5 - (1/2)(12)) * 2]
Simplifying this expression, we can calculate each term:
Riemann Sum = (5 - 1) + (5 - 2) + (5 - 3) + (5 - 4) + (5 - 5) + (5 - 6)
= 4 + 3 + 2 + 1 + 0 + (-1)
= 9
Therefore, the Riemann sum for the given function, with six subintervals using left endpoints as sample points, is 9.