smc-5145-30-Estimate comparison

Calculus, MET2 2025 VCAA 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

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

♦ Mean mark 50%.

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

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

\(\Rightarrow B\)

Calculus, 2ADV C4 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)

--- 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)

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

Show Answers Only

`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)

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

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

Show Answers Only

`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)

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

--- 4 WORK AREA LINES (style=lined) ---
Using parts (i) and (ii), show that `pi ~~ (12 sqrt 3)/(3 + 2 sqrt 3)`. (1 mark)

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

Show Answers Only

`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)

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

Show Answers Only

`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)

--- 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 (ii) 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

`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).}`

`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}`