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5 paź 2023 · Single and composite applications of the trapezoidal rule to approximate the value of definite integrals. Error analysis of the trapezoidal rule.
Example 1. Approximate the integral of f(x) = x 3 on the interval [1, 2] with four subintervals. First, h = (2 - 1)/4 = 0.25, and thus we calculate: ½⋅(f(1) + 2⋅(f(1.25) + f(1.5) + f(1.75)) + f(2))⋅0.25 = ½⋅(1 3 + 2⋅(1.25 3 + 1.5 3 + 1.75 3) + 2 3)⋅0.25 = 3.796875
For example, Figure 1 shows a single application of the trapezoidal rule. Figure 1. The integral of cos(x) on [0, π/2] approximated with one trapezoid. Figure 2 shows the trapezoidal rule applied to four sub-intervals on the interval [0, π/2].
Exercise 1: Testing the accuracy of the composite trapezoidal rule. First we have to recall the error estimate for for the trapezoidal rule on a single interval [a,b]. If f 2C2(a,b), then there is a x 2(a,b) such that I[f] T[f] = (b a)3 12 f00(x). 4.4 Theorem 1: Quadrature error estimate for composite trapezoidal rule
Transfer the formula from [ 11] to some interval [ ]. a, b. Design a composite formula, by dividing the interval [ ] into subintervals and applying the quadrature formula on each subinterval. Ef a, b I f a, b − Q f a, b. Find an expression for the error [ ]( ) = [ ]( ) [ ]( ).
Use the composite trapezoidal rule with two segments to find the distance covered by the body from \(t = 2\) to \(t = 9\) seconds. (A) \(935.0 m\) (B) \(1039.7 m\)
In this lesson we will learn to obtain high accuracy approximations to definite integrals using the Composite Trapezoidal Rule. n−1 Zb h (b − a)f′′(μ) X f(x) dx = f(a) + 2 f(xj) + f(b) 2. − h2. . 12 j=1. − a. where h = and xj = a + jh for j = 0, 1, . . . , n. n.
