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
i. using the trapezoidal rule
a. estimate the integral with ( n = 4 ) steps and find an upper bound for ( left|e_{t}
ight| ).
the estimate using the trapezoidal rule with ( n = 4 ) is 0.2545
(round to four decimal places as needed.)
the upper bound for ( left|e_{t}
ight| ) is
(round to four decimal places as needed.)

Explanation:

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

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the first - derivative \(y^\prime=f^\prime(s)=-2s^{-3}\), and the second - derivative \(y^{\prime\prime}=f^{\prime\prime}(s)=6s^{-4}=\frac{6}{s^{4}}\)

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

Since \(y = f^{\prime\prime}(s)=\frac{6}{s^{4}}\) is a decreasing function on the interval \((0,\infty)\) (because as \(s\) increases, \(y\) decreases). So on the interval \([2,4]\), the maximum value of \(|f^{\prime\prime}(s)|\) occurs at \(s = 2\). Then \(|f^{\prime\prime}(s)|=\frac{6}{s^{4}}\), and \(M=\max_{s\in[2,4]}|f^{\prime\prime}(s)|=\frac{6}{2^{4}}=\frac{6}{16}=\frac{3}{8}\)

Step3: Use the error formula for the trapezoidal rule \(|E_{T}|\leq\frac{(b - a)^{3}}{12n^{2}}M\)

We know that \(a = 2\), \(b = 4\), \(n = 4\), and \(M=\frac{3}{8}\)

$$
LATEXBLOCK0
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Answer:

\(0.0156\)