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Calculus, MET1 2017 VCAA 2

Let  `y = x log_e(3x).`

  1. Find  `(dy)/(dx)`.  (2 marks)
  2. Hence, calculate  `int_1^2 (log_e(3x) + 1) dx`. Express your answer in the form  `log_e(a)`, where `a` is a positive integer.  (2 marks)
Show Answers Only
  1. `(dy)/(dx) = log_e (3x) + 1`
  2. `log_e(12)`
Show Worked Solution

a.  `text(Using Product Rule:)`

`(hg)′` `= h′ g + h g′`
`:. (dy)/(dx)` `= 1 xx log_e (3x) + x (3/(3x))`
  `= log_e (3x) + 1`

 

b.  `text(Using Integration by Recognition:)`

MARKER’S COMMENT: Too many students ignored the hence instruction and were not able to form an integral.

`int (log_e(3x) + 1) dx = x log_e (3x)`

`:. int_1^2 (log_e (3x) + 1) dx`

`= [x log_e (3x)]_1^2`

`= (2 log_e (6)) – (log_e(3))`

`= log_e(6^2) – log_e(3)`

`= log_e (36/3)`

`= log_e(12)`

Calculus, MET1 2016 ADV 12d

  1. Differentiate  `y = xe^(3x)`.  (1 mark)
  2. Hence find the exact value of  `int_0^2 e^(3x) (3 + 9x)\ dx`.  (2 marks)
Show Answers Only
  1. `e^(3x) (1 + 3x)`
  2. `6e^6`
Show Worked Solution

a.  `y = xe^(3x)`

`text(Using product rule,)`

`(dy)/(dx)` `= x · 3e^(3x) + 1 · e^(3x)`
  `= e^(3x) (1 + 3x)`

 

b.  `int_0^2 e^(3x) (3 + 9x)\ dx`

`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6 – 0)`

`= 6e^6`

Calculus, MET2 2016 VCAA 9 MC

Given that  `(d(xe^(kx)))/(dx) = (kx + 1)e^(kx)`, then  `int xe^(kx) dx` is equal to

  1. `(xe^(kx))/(kx + 1) + c`
  2. `((kx + 1)/k)e^(kx) + c`
  3. `1/k int e^(kx) dx`
  4. `1/k (xe^(kx) - int e^(kx) dx) + c`
  5. `1/k^2 (xe^(kx) - e^(kx)) + c`
Show Answers Only

`D`

Show Worked Solution
♦ Mean mark 41%.
`int (kx + 1)e^(kx) dx` `= xe^(kx) + c_1`
`k int xe^(kx) dx + int e^(kx) dx` `= xe^(kx) + c_1`
`k int xe^(kx) dx` `= xe^(kx) – int e^(kx) dx + c_1`
`:. int xe^(kx) dx` `= 1/k (xe^(kx) – int e^(kx) dx) + c`

 
`=>   D`