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Calculus, 2ADV C4 2020 HSC 18

  1. Differentiate  `e^(2x) (2x + 1)`.  (2 marks)

  2. Hence, find  `int(x + 1)e^(2x)\ dx`.  (1 marks)

Show Answers Only
  1. `4e^(2x)(x + 1)`
  2. `1/4 e^(2x)(2x + 1) + c`
Show Worked Solution
a.    `y` `= e^(2x) (2x + 1)`
  `(dy)/(dx)` `= 2e^(2x)(2x + 1) + 2e^(2x)`
    `= 2e^(2x)(2x + 2)`
    `= 4e^(2x)(x + 1)`

♦ Mean mark part (b) 40%.

 

b.    `int(x + 1)e^(2x)dx` `= 1/4 int 4e^(2x)(x + 1)`
    `= 1/4 e^(2x)(2x + 1) + c`

Calculus, 2ADV C4 2019 HSC 13c

  1.  Differentiate  `(ln x)^2`.  (2 marks)

  2.  Hence, or otherwise, find  `int(ln x)/x\ dx`.  (1 mark)

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  1. `(2 ln x)/x`
  2. `1/2 (ln x)^2 + C`
Show Worked Solution
i.    `y` `= (ln x)^2`
  `(dy)/(dx)` `= 2 ⋅ 1/x ⋅ ln x`
    `= (2 ln x)/x`

♦ Mean mark part (ii) 49%.

ii.    `int (ln x)/x\ dx` `=1/2 int (2 ln x)/x dx`
    `= 1/2 (ln x)^2 +C`

Calculus, 2ADV C4 2016 HSC 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

i.  `y = xe^(3x)`

`text(Using product rule:)`

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

 

ii.  `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`