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Calculus, MET1 2023 VCE SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

Calculus, MET1 2013 VCAA 6

Let  `g: R -> R,\ \ g(x) = (a - x)^2`,  where `a` is a real constant.

The average value of `g` on the interval  `[– 1, 1]`  is  `31/12.`

Find all possible values of `a.`  (3 marks)

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`+- 3/2`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 39%.
MARKER’S COMMENT: “Far too many” students misunderstood “average value” in this context.
`1/(1 – (– 1)) int_-1^1 (a – x)^2 dx` `= 31/12`
`[ax^2 – ax^2 + x^3/3]_-1^1` `= 31/6`
`[(a^2 – a+1/3) – (-a^2-a-1/3)]` `=31/6`
`2a^2+2/3` `=31/6`
`a^2` `=27/4`
`:. a` `=+- 3/2`

 

`text(Solution 2)`

`1/(1 – (– 1)) int_-1^1 (a – x)^2 dx` `= 31/12`
`1/2 [(a – x)^3/-3]_-1^1` `= 31/12`
`[(a – x)^3]_-1^1` `= – 31/2`
`(a – 1)^3 – (a + 1)^3` `= – 31/2`

`(a^3 – 3a^2 + 3a – 1) – (a^3 + 3a^2 + 3a + 1)=- 31/2`

`-6a^2 – 2` `= – 31/2`
`6a^2` `= 27/2`
`a^2` `= 9/4`
`:. a` `= +- 3/2`

Calculus, MET1 2015 VCAA 4

Consider the function  `f:[−3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2 - 4)`.

  1. Find the coordinates of the stationary points of the function.  (2 marks)

The rule for  `f` can also be expressed as  `f(x) = 1/2(x - 1)(x + 2)^2`.

  1. On the axes below, sketch the graph of  `f`, clearly indicating axis intercepts and turning points.

     

    Label the end points with their coordinates.  (2 marks)

      

    met1-2015-vcaa-q4
     

  2. Find the average value of  `f` over the interval  `0 <= x <=2`.  (2 marks)
Show Answers Only
  1. `(0, − 2), ( − 2,0)`
  2.  

     

    met1-2015-vcaa-q4-answer

  3. `1`
Show Worked Solution

a.   `text(Stationary points when)\ \ f´(x)=0,`

`1/2(3x^2 + 6x)` `= 0`
`3x(x + 2)` `= 0`

 

`:. x = 0, − 2`

 

`:.\ text(Coordinates of stationary points:)`

`(0, − 2), (− 2,0)`

 

b.    met1-2015-vcaa-q4-answer
♦ Part (c) mean mark 50%.
MARKER’S COMMENT: Most students recalled the average value definition but then did not integrate correctly.

 

c.    `text(Avg value)` `= 1/(2 – 0) int_0^2 f(x) dx`
    `= 1/2 int_0^2 1/2(x^3 + 3x^2 – 4)dx`
    `= 1/4[1/4x^4 + x^3 – 4x]_0^2`
    `= 1/4[(16/4 + 2^3 – 4(2))-0]`
    `= 1`