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Calculus, MET1 2022 VCAA 2a

Let `g:\left(\frac{3}{2}, \infty\right) \rightarrow R, g(x)=\frac{3}{2 x-3}`

Find the rule for an antiderivative of `g(x)`.   (1 mark)

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`\frac{3}{2} \log _e(2 x-3)`

Show Worked Solution
`\int \frac{3}{2 x-3} d x` `=frac{3}{2}\int \frac{2}{2 x-3} d x`  
  `=\frac{3}{2} \log _e(2 x-3)`  

Notes:

→ As the domain is `\left(\frac{3}{2}, \infty\right)`  the natural log to the base `e` is used

→ A constant is not required in the solution as the question only asks for AN antiderivative not a family of solutions.


♦ Mean mark 50%.

Calculus, MET1 2018 VCAA 2

The derivative with respect to `x` of the function  `f:(1,∞) -> R`  has the rule  `f′(x) = 1/2 - 1/(2x - 2)`.

Given that  `f(2) = 0`, find  `f(x)`  in terms of `x`.  (3 marks)

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`f(x) =1/2[x – ln(x – 1)] – 1`

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`fprime(x) = 1/2 – 1/(2x – 2)`

`f(x)` `= int(1/2 – 1/(2x -2))\ dx`
  `= 1/2 int(1 – 1/(x – 1))\ dx`
  `= 1/2[x – ln|x – 1|] + c`

 
`text(Given)\ \ f(2) = 0,`

`0 = 1/2(2 – ln 1) + c`

`c = −1`

`:. f(x) = 1/2[x – ln|x – 1|] – 1`

Calculus, MET1 2006 ADV 2bi

If  `f′(x)= (x^2)/(x^3 + 1)`  and  `f(1)= log_e 2,` find  `f(x)`.  (3 marks)

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`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

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`f(x)` `=int f′(x)\ dx`
  `=int (x^2)/(x^3 + 1)\ dx`
  `= 1/3 int (3x^2)/(x^3 + 1)\ dx`
  `=1/3 log_e |(x^3 + 1)|+c`

 

`text(S)text(ince)\ \ f(1)= log_e 2,`

`1/3 log_e 2+c` `= log_e 2`
`c` `=2/3 log_e 2`
   

`:.\ f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Calculus, MET1 2015 VCAA 2

Let  `f′(x) = 1 - 3/x`, where  `x != 0`.

Given that  `f(e) = −2`, find  `f(x)`.  (3 marks)

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`x – 3log_e(x) + 1 – e`

Show Worked Solution
`f(x)` `= int 1 – 3x^(−1) dx`
`f(x)` `= x – 3log_e |\ x\ | + c`

 

`text(Substitute)\ \ f(e) = − 2,`

`−2` `= e – 3log_e(e) + c`
`c` `= 1- e`

 

`:. f(x) = x – 3log_e|\ x\ | + 1 – e`