smc-976-30-Estimate Comparison

Calculus, 2ADV C4 2025 HSC 27

The shaded region is bounded by the graph \(y=\left(\dfrac{1}{2}\right)^x\), the coordinate axes and \(x=2\).

Use two applications of the trapezoidal rule to estimate the area of the shaded region. (2 marks)

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Show that the exact area of the shaded region is \(\dfrac{3}{4 \ln 2}\). (2 marks)

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Using your answers from part (a) and part (b), deduce \(e<2 \sqrt{2}\). (2 marks)

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a. \(A\approx \dfrac{9}{8}\ \text{units}^2\)

b.	\(\text{Area}\)	\(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
		\(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
		\(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
		\(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
		\(=\dfrac{3}{4 \ln 2}\)

c. \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\)	\(>\dfrac{3}{4 \ln 2}\)
\(36\, \ln 2\)	\(>24\)
\(\ln 2\)	\(>\dfrac{2}{3}\)
\(e^{\small \dfrac{2}{3}}\)	\(>2\)
\(e\)	\(>2^{\small \dfrac{3}{2}}\)
\(e\)	\(>2 \sqrt{2}\)

Show Worked Solution

\begin{array}{|c|c|c|c|}
\hline \ \ x \ \ & \ \ 0 \ \ & \ \ 1 \ \ & \ \ 2 \ \ \\
\hline y & 1 & \dfrac{1}{2} & \dfrac{1}{4} \\
\hline
\end{array}

\(A\)	\(\approx \dfrac{h}{2}\left[1 \times 1+2 \times \dfrac{1}{2}+1 \times \dfrac{1}{4}\right]\)
	\(\approx \dfrac{1}{2}\left(\dfrac{9}{4}\right)\)
	\(\approx \dfrac{9}{8}\ \text{units}^2\)

b.	\(\text{Area}\)	\(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
		\(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
		\(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
		\(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
		\(=\dfrac{3}{4 \ln 2}\)

c. \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\)	\(>\dfrac{3}{4 \ln 2}\)
\(36\, \ln 2\)	\(>24\)
\(\ln 2\)	\(>\dfrac{2}{3}\)
\(e^{\small \dfrac{2}{3}}\)	\(>2\)
\(e\)	\(>2^{\small \dfrac{3}{2}}\)
\(e\)	\(>2 \sqrt{2}\)

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 2022 HSC 29

The diagram shows the graph of `y=2^{-x}`. Also shown on the diagram are the first 5 of an infinite number of rectangular strips of width 1 unit and height `y=2^{-x}` for non-negative integer values of `x`. For example, the second rectangle shown has width 1 and height `(1)/(2)`.

The sum of the areas of the rectangles forms a geometric series.
Show that the limiting sum of this series is 2. (1 mark)

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Show that `int_(0)^(4)2^(-x)\ dx=(15)/(16 ln 2)`. (2 marks)

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Use parts (a) and (b) to show that `e^(15) < 2^(32)`. (2 marks)

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`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

a. `text{Consider the rectangle heights:}`

`2^0=1, \ 2^(-1)=1/2, \ 2^(-2)= 1/4, \ 2^(-3)= 1/8, …`

`=>\ text{Rectangle Areas}\ = 1, \ 1/2, \ 1/4, \ 1/8, …`

`a=1,\ \ r=1/2`

`S_oo=a/(1-r)=1/(1-1/2)=2\ \ text{… as required}`

b. `text{Show}\ \ int_0^4 2^(-x)\ dx = 15/(16ln2)`

`int_0^4 2^(-x)\ dx `	`=(-1)/ln2[2^(-x)]_0^4`
	`=(-1)/ln2(1/16-1)`
	`=1/ln2-1/(16ln2)`
	`=(16-1)/(16ln2)`
	`=15/(16ln2)\ \ text{… as required}`

Mean mark (b) 56%.

c. `text{Show}\ \ e^15<2^32`

`text{Area under curve < Sum of rectangle areas}`

`15/(16ln2)`	`<2`
`15`	`<32ln2`
`15/32`	`<ln2`
`e^(15/32)`	`<e^(ln2)`
`root(32)(e^15)`	`<2`
`e^15`	`<2^32\ \ text{… as required}`

♦♦♦ Mean mark (c) 9%.

Calculus, 2ADV C4 2013* HSC 15a

The diagram shows the front of a tent supported by three vertical poles. The poles are 1.2 m apart. The height of each outer pole is 1.5 m, and the height of the middle pole is 1.8 m. The roof hangs between the poles.

The front of the tent has area `A\ text(m²)`.

Use the trapezoidal rule to estimate `A`. (1 mark)

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Does the Trapezoidal rule give a higher or lower estimate of the actual area? Justify your answer. (1 mark)

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`3.96\ text(m²)`
`text(See Worked Solutions)`

Show Worked Solution

i.	`A`	`~~ h/2 [y_0 + 2y_1 + y_2]`
		`~~ 1.2/2 [1.5 + (2 xx 1.8) + 1.5]`
		`~~ 0.6 [6.6]`
		`~~ 3.96\ text(m²)`

ii.

`text(S)text(ince the tent roof is concave up, the)`

`text(Trapezoidal rule uses straight lines and)`

`text(will estimate a higher area.)`

Calculus, 2ADV C4 2004* HSC 10a

Use the Trapezoidal rule with 3 function values to find an approximation to the area under the curve `y = 1/x` between `x = a ` and `x = 3a`, where `a` is positive. (2 marks)

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Using the result in part (i), show that `ln 3 ≑ 7/6`. (1 mark)

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`7/6`
`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`A`	`~~ a/2[1/a + 2(1/(2a)) + 1/(3a)]`
	`~~ a/2(7/(3a))`
	`~~ 7/6`

ii. `text{Area under the curve}\ \ y=1/x`

`= int_a^(3a) 1/x\ dx`

`= [ln x]_(\ a)^(3a)`

`= ln 3a − ln a`

`= ln\ (3a)/a`

`= ln 3`

`text{Trapezoidal rule in part (i) found the approximate}`

`text(value of the same area.)`

`:. ln 3 ≑ 7/6.`

Calculus, 2ADV C4 2017* HSC 14b

Find the exact value of `int_0^(pi/3) cos x\ dx`. (1 mark)

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Using the Trapezoidal rule with three function values, find an approximation to the integral `int_0^(pi/3) cos x\ dx,` leaving your answer in terms of `pi` and `sqrt 3`. (2 marks)

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Using parts (i) and (ii), show that `pi ~~ (12 sqrt 3)/(3 + 2 sqrt 3)`. (1 mark)

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`sqrt 3/2`
`((2sqrt3 + 3)pi)/24`
`text{Proof (See Worked Solutions)}`

Show Worked Solution

i.	`int_0^(pi/3) cos x\ dx`	`= [sin x]_0^(pi/3)`
		`= sin\ pi/3-0`
		`= sqrt 3/2`

ii.

\begin{array} {|l|c|c|c|}\hline
x & \ \ \ 0\ \ \ & \ \ \ \dfrac{\pi}{6}\ \ \ & \ \ \ \dfrac{\pi}{3}\ \ \ \\ \hline
\text{height} & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \hline
\text{weight} & 1 & 2 & 1 \\ \hline \end{array}

`int_0^(pi/3) cos x\ dx`	`~~ 1/2 xx pi/6[1 + 2(sqrt3/2) + 1/2]`
	`~~ pi/12((3 + 2sqrt3)/2)`
	`~~ ((3+2sqrt3)pi)/24`

♦ Mean mark part (iii) 49%.

(iii)	`((3+2sqrt3)pi)/24`	`~~ sqrt3/2`
	`:. pi`	`~~ (24sqrt3)/(2(3+2sqrt3))`
		`~~ (12sqrt3)/(3 + 2sqrt3)`

Calculus, 2ADV C4 2016* HSC 14a

The diagram shows the cross-section of a tunnel and a proposed enlargement.

The heights, in metres, of the existing section at 1 metre intervals are shown in Table `A.`

The heights, in metres, of the proposed enlargement are shown in Table `B.`

Use the Trapezoidal rule with the measurements given to calculate the approximate increase in area. (3 marks)

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`1.3\ text(m²)`

Show Worked Solution

`text(Consider the shaded area distances:)`

`A`	`~~ 1/2[0 + 2(0.4 + 0.5 + 0.4) + 0]`
	`~~ 1/2(2.6)`
	`~~ 1.3\ text(m²)`

Calculus, 2ADV C4 2010 HSC 3b

Sketch the curve `y=lnx`. (1 mark)

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Use the trapezoidal rule with 3 function values to find an approximation to `int_1^3 lnx\ dx` (2 marks)

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State whether the approximation found in part (ii) is greater than or less than the exact value of `int_1^3 lnx\ dx`. Justify your answer. (1 mark)

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`text(See Worked Solutions for sketch.)`
`1.24\ \ text(u²)`
`text(See Worked Solutions)`

Show Worked Solutions

MARKER’S COMMENT: Many students failed to illustrate important features in their graph such as the concavity, `x`-axis intercept and `y`-axis asymptote (this can be explicitly stated or made graphically clear).

ii. `text(Area)`	`~~h/2[f(1)+2xxf(2)+f(3)]`
	`~~1/2[0+2ln2+ln3]`
	`~~1/2[ln(2^2 xx3)]`
	`~~1/2ln12`
	`~~1.24\ \ text(u²)` `text{(to 2 d.p.)}`

iii.

♦♦♦ Mean mark 12%.
MARKER’S COMMENT: Best responses commented on concavity and that the trapezia lay beneath the curve. Diagrams featured in the best responses.

`text(The approximation is less because the sides)`

`text{of the trapezia lie below the concave down}`

`text{curve (see diagram).}`