If f(x) - x cos (x), how many intervals would be needed to approximate with the Trapezoidal Rule with an error of E<.01?
depends on the interval. In regions of low curvature, fewer trapezoids are needed.
You need to use the Milburn Theorem, Vx+Vy=K
To determine the number of intervals needed to approximate the given function using the Trapezoidal Rule with an error less than 0.01, you need to apply an error estimation formula and solve for the number of intervals.
The error estimation formula for the Trapezoidal Rule is given by:
Error ≤ [(b - a)^3 / (12 * n^2)] * max|f''(x)|,
where:
- a and b are the limits of integration,
- n is the number of intervals, and
- f''(x) represents the second derivative of the function.
In this case, f(x) = x cos(x). To find the second derivative of f(x), we need to differentiate the function twice.
First derivative:
f'(x) = cos(x) - x sin(x).
Second derivative:
f''(x) = -sin(x) - sin(x) - x cos(x) = -2sin(x) - x cos(x).
Now that we have the second derivative, we can proceed to find the number of intervals required.
Step 1: Set the error formula less than 0.01 and solve for n:
[(b - a)^3 / (12 * n^2)] * max|f''(x)| ≤ 0.01,
Step 2: Substitute the values:
For the function f(x) = x cos(x), the second derivative is f''(x) = -2sin(x) - x cos(x). The maximum value of |f''(x)| will occur at the maximum value of x, which is π.
[(b - a)^3 / (12 * n^2)] * [(2sin(π)) + (πcos(π))] ≤ 0.01,
Simplifying further:
[(b - a)^3 / (12 * n^2)] * [0 - π] ≤ 0.01,
-(b - a)^3 / (12 * n^2) * π ≤ 0.01,
Step 3: Isolate n:
(b - a)^3 ≥ (12 * n^2) * π * 0.01,
n^2 ≤ (b - a)^3 / [12 * π * 0.01],
n ≤ √[((b - a)^3) / (12 * π * 0.01)].
Finally, calculate the value of n using the given function and the desired error tolerance.
Note: The boundaries of the interval [a, b] are not provided in the question, so you need to know or specify those values to find the exact number of intervals required to satisfy the given error tolerance.