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Calculus, 2ADV C4 2024 HSC 22

The graph of the function  \(f(x) = \ln(1 + x^{2})\)  is shown.
 

  1. Prove that \(f(x)\) is concave up for  \(-1 < x < 1\).   (3 marks)
  2. A table of function values, correct to 4 decimal places, for some \(x\) values is provided.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}

  1. Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram.   (2 marks)
  2. Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer.   (1 mark)
Show Answers Only

a.    \(f(x)= \ln(1+x^2)\)

\(f^{\prime}(x)=\dfrac{2x}{1+x^2}\)

\(f^{\prime\prime}(x)\) \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)
  \(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)
  \(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)

 
\(\text{Consider domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{\prime\prime}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)
 

b.    \(\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}\)

c.    \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal}\)

\(\text{rule will overestimate the area.}\)

Show Worked Solution

a.    \(f(x)= \ln(1+x^2)\)

\(f^{′}(x)=\dfrac{2x}{1+x^2}\)

\(f^{″}(x)\) \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)  
  \(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)  
  \(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)  

 
\(\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)

♦ Mean mark (a) 47%.

b.    \(\text{Total shaded area}\)

\(\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] \)

\(\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] \)

\(\approx 0.538275 \)

\(\approx 0.5383\ \text{(4 d.p.)}\)

♦ Mean mark (b) 50%.

c.    \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal}\)

\(\text{rule will overestimate the area.}\)

♦ Mean mark (c) 49%.