smc-976-10-Table provided

Calculus, 2ADV C4 2024 HSC 22

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

Prove that \(f(x)\) is concave up for \(-1 < x < 1\). (3 marks)
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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}

Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram. (2 marks)
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Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer. (1 mark)
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a. \(f(x)= \ln(1+x^2)\)

\(f^{′}(x)=\dfrac{2x}{1+x^2}\)

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

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

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

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

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

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

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

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

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

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

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

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

♦ Mean mark (c) 49%.

Calculus, 2ADV C4 2020 HSC 20

Kenzo is driving his car along a road while his friend records the velocity of the car, `v(t)`, in km/h every minute over a 5-minute period. The table gives the velocity `v(t)` at time `t` hours.

The distance covered by the car over the 5-minute period is given by

`int_0^(5/60) v(t)\ dt`.

Use the trapezoidal rule and the velocity at each of the six time values to find the approximate distance in kilometres the car has travelled in the 5-minute period. Give your answer correct to one decimal place. (2 marks)

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`5.4\ text(km)`

Show Worked Solution

`int_0^(5/60) v(t)\ dt`	`~~ 1/2 xx 1/60 [60 + 2(55 + 65 + 68 + 70) + 67]`
	`~~ 1/120 (643)`
	`~~ 5.358…`
	`~~ 5.4\ text(km)`

Calculus, 2ADV C4 2011* HSC 5c

The table gives the speed `v` of a jogger at time `t` in minutes over a 20-minute period. The speed `v` is measured in metres per minute, in intervals of 5 minutes.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ t\ \ \ \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \ &\ \ \ 5\ \ \ &\ \ \ 10\ \ \ &\ \ \ 15\ \ \ &\ \ \ 20\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ \ v\ \ \ \rule[-1ex]{0pt}{0pt} & 173 & 81 & 127 & 195 & 168 \\
\hline
\end{array}

The distance covered by the jogger over the 20-minute period is given by `int_0^20 v\ dt`.

Use the Trapezoidal rule and the speed at each of the five time values to find the approximate distance the jogger covers in the 20-minute period. (3 marks)

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`text(2867.5 metres)`

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