Express the limit as a definite integral on the given interval. Use Riemright and evaluate the sum by hand.
lim n-> infinity sigma with n on top and i=1 on the bottom [4 - 3(xi*)^2 + 6(xi*)^5]delta x, [0,2]
Please help. I am really confused. To find delta x I would use (b-a)/n , but I am not given any of those values. And how do I know how many rectangles and where to break it up?
You are given b and a
b = 2
a = 0
However you are not given the number of rectangles, n
Perhaps you can choose your own number.
The more rectangles you use, the closer the approximation will be to the actual integral
As n-->infinity, the answer becomes exact.
integral from 0 to 2 of (4-3x^2+6 x^5) dx
The exact answer is
(4x -x^3 +x^6) at 2 - at 0
or
8-8+64 = 64
see how close to 64 you come with n = 3
n = 3 etc
To express the limit as a definite integral on the given interval, we need to first understand the limit as a Riemann sum and then convert it into a definite integral. Here's a step-by-step guide:
1. Recall the definition of a Riemann sum for a function f(x) on an interval [a, b] with n subintervals:
R = ∑[i=1 to n] f(xi*)Δx
Here, xi* represents a sample point in each subinterval, and Δx is the width of each subinterval.
2. In this case, the given limit is:
lim[n→∞] ∑[i=1 to n] [4 - 3(xi*)^2 + 6(xi*)^5]Δx, on the interval [0,2]
3. To express this limit as a definite integral, we can recognize that the term ∑[i=1 to n] [4 - 3(xi*)^2 + 6(xi*)^5]Δx is essentially a Riemann sum. We can convert it to a definite integral by taking the limit as n approaches infinity.
The definite integral form is:
∫[0 to 2] [4 - 3x^2 + 6x^5] dx
4. To evaluate the sum by hand using the Riemann sum, we need to determine Δx and the sample points xi* for the given interval.
To find Δx, we can use the formula Δx = (b - a)/n. In this case, a = 0 and b = 2. However, the value of n is not given explicitly, so we'll have to make an assumption based on the problem context or use a general approach.
Let's assume n is a positive integer. If we choose n = 2 for simplicity, then Δx = (2 - 0)/2 = 1.
Now, we need to determine the sample points xi* (also known as the partition points). For n = 2, we'll have three partition points: x1*, x2*, and x3*.
Assuming equal subintervals, we have:
x1* = 0 + Δx/2 = 0 + 1/2 = 0.5
x2* = x1* + Δx = 0.5 + 1 = 1.5
x3* = x2* + Δx = 1.5 + 1 = 2.5
5. Now, we can substitute these values into the Riemann sum:
R = [4 - 3(0.5)^2 + 6(0.5)^5]Δx + [4 - 3(1.5)^2 + 6(1.5)^5]Δx
Calculating each term and simplifying:
R = [4 - 0.75 + 0.015] * 1 + [4 - 6.75 + 56.25] * 1
R = 3.265 + 53.5
R ≈ 56.765
6. Therefore, the approximated value of the limit using Riemann sum is 56.765.
Note: It's important to mention that the Riemann sum approximation becomes more accurate as the number of subintervals (n) increases and approaches infinity.