PaulKraemer
New member
- Joined
- Apr 10, 2011
- Messages
- 45
Hi,
I am having trouble with the following problem that asks you to find the area under the graph of f = sqrt (9 - x^2) from 0 to 3. This problem comes in a chapter before you learn to evaluate integrals. I am following the method used to solve the examples in this chapter, which is basically as follows:
1. Divide the interval [0,3] into n equal subintervals. This makes the length of typical subinterval delta-x = 3/n
x0 = 0, x1= delta-x, x2= 2 * delta-x, xi = i * delta-x, ... , xn = n * delta-x = 3
Because delta-x = 3/n, I get...
xi = i * delta-x = 3i/n
2. Since f is decreasing on [0,3], the number ui in [xi-1, xi] at which f takes on its minimum value is always the right-hand endpoint, that is:
ui = xi = 3i/n
3. f(ui) = f(3i/n) = sqrt(9 - (3i/n)^2)
4. f(ui) * delta-x = sqrt(9 - (3i/n)^2) * 3/n.
...this is where I get stuck. In all of the examples in this chapter, they solve for the area by calculating the limit as n approaches infinity of the sum from 1 to n of f(ui) * delta-x.
In all the examples, they simplify this sum by reducing it to sums of i, i^2, or i^3, where they give you the formulas for these sums in terms of n.
None of the examples had the i inside a sqrt. I can't figure out how to reduce the formula for the sum 1 to n of f(ui) * delta-x to anything I know how to deal with.
Any help will be greatly appreciated. Thanks in advance.
Paul
I am having trouble with the following problem that asks you to find the area under the graph of f = sqrt (9 - x^2) from 0 to 3. This problem comes in a chapter before you learn to evaluate integrals. I am following the method used to solve the examples in this chapter, which is basically as follows:
1. Divide the interval [0,3] into n equal subintervals. This makes the length of typical subinterval delta-x = 3/n
x0 = 0, x1= delta-x, x2= 2 * delta-x, xi = i * delta-x, ... , xn = n * delta-x = 3
Because delta-x = 3/n, I get...
xi = i * delta-x = 3i/n
2. Since f is decreasing on [0,3], the number ui in [xi-1, xi] at which f takes on its minimum value is always the right-hand endpoint, that is:
ui = xi = 3i/n
3. f(ui) = f(3i/n) = sqrt(9 - (3i/n)^2)
4. f(ui) * delta-x = sqrt(9 - (3i/n)^2) * 3/n.
...this is where I get stuck. In all of the examples in this chapter, they solve for the area by calculating the limit as n approaches infinity of the sum from 1 to n of f(ui) * delta-x.
In all the examples, they simplify this sum by reducing it to sums of i, i^2, or i^3, where they give you the formulas for these sums in terms of n.
None of the examples had the i inside a sqrt. I can't figure out how to reduce the formula for the sum 1 to n of f(ui) * delta-x to anything I know how to deal with.
Any help will be greatly appreciated. Thanks in advance.
Paul