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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)

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  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`

Calculus, 2ADV C4 2009 HSC 2b

  1. Find  `int 5\ dx`.   (1 mark)

  2. Find  `int 3/((x - 6)^2)\ dx`.    (2 marks)

  3. Evaluate  `int_1^4 x^2 + sqrtx\ dx`.   (3 marks)

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  1. `5x + C`
  2. `(-3)/((x – 6)) + C`
  3. `77/3`
Show Worked Solution

i.  `int 5\ dx= 5x + C`

 

ii.  `int 3/((x – 6)^2)\ dx`

`= 3 int (x – 6)^(-2)\ dx`

`= 3 xx 1/(-1) xx (x – 6)^(-1) + c`

`= (-3)/((x – 6)) + c`

 

iii.  `int_1^4 x^2 + sqrtx\ \ dx`

`= int_1^4 (x^2 + x^(1/2))\ dx`

`= [1/3 x^3 + 1/(3/2) x^(3/2)]_1^4`

`= [(x^3)/3 + 2/3x^(3/2)]_1^4`

`= [((4^3)/3 + 2/3 xx 4^(3/2))\ – (1/3 + 2/3)]`

`= [(64/3 + 16/3) – 3/3]`

`= [80/3 – 3/3]`

`= 77/3`

Calculus, 2ADV C4 2010 HSC 2e

Given that  `int_0^6 ( x + k )\ dx = 30`, and  `k`  is a constant, find the value of  `k`.   (2 marks)

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`k = 2`

Show Worked Solution
`int_0^6 ( x + k ) \ dx` `= 30`
`int_0^6 ( x + k ) \ dx` `= [ 1/2\ x^2 + kx ]_0^6`
  `= [(1/2xx 6^2 + 6 xx k ) – 0 ]`
  `= 18 + 6k`

 

`=> 18 + 6k` `=30`
`6k` `= 12`
`:.  k` `= 2`