smc-7132-30-Estimate vs Actual

Calculus, 2ADV C4 2025 MET2 6 MC

The trapezium rule is used, with two trapeziums, to estimate the area bounded by the graph of \(y=f(x)\), the \(x\)-axis and the lines \(x=0\) and \(x=1\).

For which function will the trapezium rule estimate be larger than the exact area?

\(f(x)=3-e^x\)
\(f(x)=x^3+1\)
\(f(x)=3 \sin (x)+1\)
\(f(x)=\log _e(x+3)\)

Show Answers Only

\(B\)

Show Worked Solution

\(\text{Rough sketch the shape of each graph.}\)

\(\text{Consider option B:}\)

♦ Mean mark 50%.

\(\text{By drawing trapeziums (see graph), the estimated area}\)

\(\text{is greater than the actual area.}\)

\(\Rightarrow B\)

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)

--- 5 WORK AREA LINES (style=lined) ---
Show that the exact area of the shaded region is \(\dfrac{3}{4 \ln 2}\). (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Using your answers from part (a) and part (b), deduce \(e<2 \sqrt{2}\). (2 marks)

--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Mean mark (b) 51%.

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

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

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

♦♦♦ Mean mark (c) 25%.

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

Calculus, 2ADV C4 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(m)^2`.

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

--- 2 WORK AREA LINES (style=lined) ---
Does the Trapezoidal rule give a higher or lower estimate of the actual area? Justify your answer. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `3.96\ text(m)^2`

b. `text(See Worked Solutions)`

Show Worked Solution

a.	`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(m)^2`

`text(The tent roof is concave up. Since the Trapezoidal rule uses straight lines,)`

`text(it 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)

--- 4 WORK AREA LINES (style=lined) ---
Using the result in part (a), show that `ln 3 ≑ 7/6`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `7/6`

b. `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`

b. `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 (a) found the approximate value of the same area.)`

`:. ln 3 ≑ 7/6.`

Calculus, 2ADV C4 2010 HSC 3b

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

--- 6 WORK AREA LINES (style=lined) ---
Use the trapezoidal rule with 3 function values to find an approximation to `int_1^3 lnx\ dx` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
State whether the approximation found in part (b) is greater than or less than the exact value of `int_1^3 lnx\ dx`. Justify your answer. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answer Only

a. `text(See Worked Solutions for sketch.)`

b. `1.24\ text(u)^2`

c. `text(See Worked Solutions)`

Show Worked Solutions

MARKER’S COMMENT: Important features of the graph should be identified (as shown).

b. `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{u}^2\ \text{(to 2 d.p.)}`

♦♦♦ Mean mark 12%.
MARKER’S COMMENT: Best responses commented on concavity, trapezia laying under the curve and featured diagrams.

`text{The approximation is less because the sides of the trapezia lie}`

`text{below the concave down curve (see diagram).}`

`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}`

`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}`