Question

the instructions for the given integral have two parts, one for the trapezoidal rule and one for simpsons rule. complete the following parts.
\\( \int _ { 2 } ^ { 4 } \frac { 1 } { s ^ { 2 } } d s \\)
(simplify your answer.)
c. use the formula \\( ( | e _ { t } | / ( \text { true value } ) ) \times 100 \\) to express \\( | e _ { t } | \\) as a percentage of the integrals true value.
\\( | e _ { t } | \\) as a percentage of the integrals true value is \\( 1.80 \\% \\).
(round to two decimal places as needed.)
ii. using simpsons rule
a. estimate the integral with \\( n = 4 \\) steps and find an upper bound for \\( | e _ { s } | \\).
the estimate using simpsons rule with \\( n = 4 \\) is \\( 0.2502 \\).
(round to four decimal places as needed.)
the upper bound for \\( | e _ { s } | \\) is \\( \square \\).
(round to four decimal places as needed.)

Explanation:

Step1: Find the fourth - derivative of \(f(s)=\frac{1}{s^{2}}=s^{- 2}\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\):
\(f^\prime(s)=-2s^{-3}\)
\(f^{\prime\prime}(s)=6s^{-4}\)
\(f^{\prime\prime\prime}(s)=-24s^{-5}\)
\(f^{(4)}(s)=120s^{-6}=\frac{120}{s^{6}}\)

Step2: Determine the maximum of \(|f^{(4)}(s)|\) on \([2,4]\)

Since \(y = f^{(4)}(s)=\frac{120}{s^{6}}\) is a decreasing function on the interval \([2,4]\) (because the exponent of \(s\) is negative), the maximum value of \(|f^{(4)}(s)|\) occurs at \(s = 2\).
\(M=\max_{s\in[2,4]}|f^{(4)}(s)|=\frac{120}{2^{6}}=\frac{120}{64}=\frac{15}{8}\)

Step3: Use the error formula for Simpson's rule \(|E_S|\leq\frac{(b - a)^5}{180n^4}M\)

Here, \(a = 2\), \(b = 4\), \(n = 4\), and \(M=\frac{15}{8}\)
First, calculate \((b - a)\): \(b - a=4 - 2 = 2\)
Then, substitute into the formula:

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Answer:

\(0.0013\)