Question

Find the minimum value of n that guarantees an error of no more than 1/30,000 in approximating integral 1/x dx from 3 to 4 by the Trapezoidal Rule with n equal subintervals.

Answer 1

To find the minimum value of n that guarantees an error of no more than 1/30000 in approximating integral 1/x dx from 3 to 4 by the trapezoidal rule with n so the answer to your question is N=71 and i hope you understand my answer.

Answer 2

We have that for the Question "Find the minimum value of n that guarantees an error of no more than 1/30,000 in approximating integral 1/x dx from 3 to 4 by the Trapezoidal Rule with n equal sub intervals" it can be said that  the minimum value of n that guarantees an error of no more than 1/30,000 is

$n \geq 9$

From the question we are told

Find the minimum value of n that guarantees an error of no more than 1/30,000 in approximating integral 1/x dx from 3 to 4 by the Trapezoidal Rule with n equal sub intervals

Generally the equation for the  minimum value   is mathematically given as

$E=\frac{K(b-a^3)}{12n^2}\\\\Here\\\\f''(x) \leq K$

Therefore

$f(x)=\frac{1}{x}\\\\f''(x)=\frac{2}{x^3}\\\\|\frac{2}{x^3}|= 1/32$

Therefore

$E=\frac{1/32*1^3}{12n^2}\\\\\frac{1/32*1^3}{12n^2}=\frac{1}{30000}$

$n \geq 9$

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