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Calculus, MET2 2023 VCE SM-Bank 5 MC

The algorithm below, described in pseudocode, estimates the value of a definite integral using the trapezium rule.
 

Consider the algorithm implemented with the following inputs.

The value of the variable sum after one iteration of the while loop would be closest to

  1. 1.281
  2. 1.289
  3. 1.463
  4. 1.617
  5. 2.136
Show Answers Only

\(C\)

Show Worked Solution

\((f(x), a, b, n)\ \rightarrow\ (\log_{e}x, 1, 3, 10) \)

\(h=\dfrac{b-a}{n}=\dfrac{3-1}{10}=\dfrac{1}{5}\)

\(\text{Sum}=f(a)+f(b)=\log_{e}1+\log_{e}3=\log_{e}3\)
  

\(\text{1st iteration of}\ \textbf{while }\text{loop:}\)

\(x\) \(=a+h=1+\dfrac{1}{5}=\dfrac{6}{5}\)
\(\text{Sum}\) \(=\text{Sum}+2\times f(x)\)
  \(=\log_{e}3+2\times f\Bigg(\dfrac{6}{5}\Bigg)\)
  \(=\log_{e}3+2\times \log_{e}{\Bigg(\dfrac{6}{5}\Bigg)}\)
  \(\approx 1.46325\dots\)

 

\(\Rightarrow C\)

Filed Under: Pseudocode, Trapezium Rule and Newton Tagged With: Band 5, smc-5145-04-Trapezium rule, smc-5145-60-Pseudocode, smc-5196-10-Trapezium rule

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