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Calculus, MET2 2024 VCAA 20 MC

The function \(f: R \rightarrow R\) has an average value \(k\) on the interval \([0,2]\) and satisfies \(f(x)=f(x+2)\) for all \(x \in R\). The value of the definite integral \( {\displaystyle \int_2^6 f(x) d x } \) is

  1. \(2k\)
  2. \(3k\)
  3. \(4k\)
  4. \(6k\)
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\(C\)

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\(\text{The function }f(x)\ \text{is a periodic function}\ \rightarrow\ \text{Period}=2\)

\(\text{Average value}\ \rightarrow\ k\) \(=\dfrac{1}{2}\displaystyle\int_0^2 f(x)\, dx\)
\(2k\) \(=\displaystyle\int_0^2 f(x)\, dx\)

\(\text{As function is periodic the average value remains the same}\ \rightarrow\ k,\text{for each period.}\)

\(\therefore\ \displaystyle\int_2^6 f(x)\, dx\) \(=2\times\displaystyle\int_0^2 f(x)\, dx\)
  \(=2\times 2k=4k\)

  
\(\Rightarrow C\)

Calculus, MET2 2020 VCAA 15 MC

Part of the graph of a function  `f`, where `a>0`, is shown below.
 

The average value of  the function  `f` over the interval  `[2a, a]` is

  1. `0`
  2. `a/3`
  3. `a/2`
  4. `3a/4`
  5. `a`
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`B`

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♦♦ Mean mark 32%.

`text(Average Value)`

`=(1)/(a-(-2a))int_(-2a)^(a)f(x)\ dx`

`=(1)/(3a)(int_(-2a)^(0)(-(3)/(2)x-a)\ dx+int_(0)^(a)(2x-a)\ dx)`

`=(a)/(3)`

`=>B`

Calculus, MET2 2009 VCAA 18 MC

The average value of the function  `f: R\ text(\){text(−)1/2} -> R,\ f(x) = 1/(2x + 1)`  over the interval  `[0, k]`  is  `1/6 log_e (7).`

The value of `k` is

  1. `(-6)/(log_e(7)) - 1/2`
  2. `3`
  3. `e^3`
  4. `(-log_e(7))/(2(log_e(7) + 6))`
  5. `171`
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`B`

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`text(Solve:)\ \ 1/(k-0) int_0^k 1/(2x + 1)\ dx` `= 1/6 log_e (7)`
`text(for)\ \ k` `> 0`

`:. k = 3`

`=>   B`