smc-756-35-Other functions

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

\(2k\)
\(3k\)
\(4k\)
\(6k\)

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\(C\)

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

\(\dfrac{1}{2}\displaystyle\int_0^2 f(x)\, dx = k\ \ \Rightarrow\ \ \displaystyle\int_0^2 f(x)\, dx = 2k\)

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

\(\therefore\ \displaystyle\int_2^6 f(x)\, dx\)

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

\(\Rightarrow C\)

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

`0`
`a/3`
`a/2`
`3a/4`
`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

`(-6)/(log_e(7)) - 1/2`
`3`
`e^3`
`(-log_e(7))/(2(log_e(7) + 6))`
`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`