Trapezoidal rule error bounds problem

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The discussion focuses on applying the trapezoidal rule to estimate the integral of x^2 from 0 to 2 using four subintervals. The initial calculation yielded an estimate of 1.9375, but it was pointed out that the user incorrectly included five terms instead of four in their trapezoidal sum. The error bound was calculated as 0.0833, but the actual error of 0.7292 exceeded this bound, indicating a mistake in the computation of terms. Corrections were suggested for the individual terms in the trapezoidal sum to align with the proper definition. The user acknowledged the errors and expressed gratitude for the assistance.
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Homework Statement



a. Use the trapezoidal rule with n = 4 subintervals to estimate \int_0^2 x^2 dx.

b. Use the error bound to find the bound for the error.

c. Compute the integral exactly.

d. Verify the error is no more than the error bound.

Homework Equations



Here, based on the problem, a = 0, b = 2, and N = 4.

T_N = \frac{1}{2} Δx(f(x_0) + 2f(x_1) + ... + 2f(x_{N - 1}) + f(x_N)) where Δx = \frac{b - a}{N} and x_i = a + iΔx.

Error(T_N) \leq \frac{K_2(b - a)^3}{12N^2} where f''(x) \leq K_2 for x \in [a, b].

Error(T_N) = |T_N - \int_a^b f(x) dx|.

The Attempt at a Solution



a. I calculated Δx = \frac{2 - 0}{4} = \frac{1}{2}. Therefore, T_N = \frac{1}{2}*\frac{1}{2}(0 + \frac{1}{2} + 1 + 2 \frac{1}{4} + 4) = 1.9375.

b. Since f''(x) = 2, I used K_2 = 2. So therefore, Error(T_N) \leq \frac{2*(2)^3}{12(4)^2} = \frac{1}{12} = .0833.

c. \int_0^2 x^2 dx = 2.6667.

d. Error(T_N) = |1.9375 - 2.6667| = .7292, which is definitely outside the error bound.

What did I do wrong?

Thanks for any help, I really appreciate it.
 
Last edited:
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mrwall-e said:
a. I calculated Δx = \frac{2 - 0}{4} = \frac{1}{2}. Therefore, T_N = \frac{1}{2}*\frac{1}{2}(0 + \frac{1}{2} + 1 + 2 \frac{1}{4} + 4) = 1.9375.

You have five terms here, but by your definition of the trapezoid sum there can only be four terms in the sum, which indicates that you have incorrectly applied the definition. You used x = 0 as the first term but the first term in the sum is n = 1, not 0.
 
Steely Dan said:
You have five terms here, but by your definition of the trapezoid sum there can only be four terms in the sum, which indicates that you have incorrectly applied the definition. You used x = 0 as the first term but the first term in the sum is n = 1, not 0.

Sorry, the equation was wrong. It should be fixed now.
 
mrwall-e said:
Sorry, the equation was wrong. It should be fixed now.

That is fine, but you've still computed the individual terms incorrectly. In particular, the third and fourth terms are incorrect.
 
Steely Dan said:
That is fine, but you've still computed the individual terms incorrectly. In particular, the third and fourth terms are incorrect.

That would do it.. such a stupid mistake, thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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