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?

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

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\(B\)

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

♦ 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(m)^2`.

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

b. `text(See Worked Solutions)`

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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)`

`text(straight lines, 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)

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

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a. `7/6`

b. `text(Proof)\ \ text{(See Worked Solutions)}`

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`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 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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i. `sqrt 3/2`

ii. `((2sqrt3 + 3)pi)/24`

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

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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(m)^2`

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`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(m)^2`

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 (b) is greater than or less than the exact value of `int_1^3 lnx\ dx`. Justify your answer. (1 mark)

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

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

c. `text(See Worked Solutions)`

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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 (c) 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}`

`text{lie 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}`