Approximate
∫_-2^3(x+3)dx
via the Riemann sum. Use the partition of five subintervals (of equal length), with the sample point barx_i being the right end point of the i-th interval.
I attempted this problem and got -40/3, but it's not being accepted. I'm not sure what I'm doing wrong.
what is the interval?
so, how did you do it?
If the interval is [-2,3] then each interval has length 1, so I fail to see how you got the fraction.
To approximate the value of the integral ∫_-2^3(x+3)dx using the Riemann sum, you need to follow a few steps:
1. Start by partitioning the interval [-2, 3] into five subintervals of equal length. The length of each subinterval is given by:
Δx = (b - a) / n,
where n is the number of subintervals and a and b are the limits of integration (-2 and 3 in this case). Thus, Δx = (3 - (-2)) / 5 = 1.
2. Next, determine the right endpoint of each subinterval. Since you are using the right end point of each subinterval as the sample point (barx_i), the values for the five right endpoints will be:
x_1 = -2 + Δx = -2 + 1 = -1,
x_2 = -1 + Δx = -1 + 1 = 0,
x_3 = 0 + Δx = 0 + 1 = 1,
x_4 = 1 + Δx = 1 + 1 = 2,
x_5 = 2 + Δx = 2 + 1 = 3.
3. Now, calculate the Riemann sum using the formula:
Riemann sum = Δx * [f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5)],
where f is the function inside the integral, in this case, (x+3).
Plugging in the values, the Riemann sum becomes:
Riemann sum = (1) * [(f(-1)) + (f(0)) + (f(1)) + (f(2)) + (f(3))].
4. Evaluate f(x) = x + 3 for each value of x:
f(-1) = -1 + 3 = 2,
f(0) = 0 + 3 = 3,
f(1) = 1 + 3 = 4,
f(2) = 2 + 3 = 5,
f(3) = 3 + 3 = 6.
5. Substitute these values back into the Riemann sum formula:
Riemann sum = (1) * [2 + 3 + 4 + 5 + 6] = 20.
Thus, the approximate value of the integral ∫_-2^3(x+3)dx using the Riemann sum with the given partition and right endpoints is 20.
It seems that there might be an error in your calculation, as you obtained -40/3 instead of 20. Please double-check your calculations and ensure that you followed the steps correctly.