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Calculus, 2ADV C4 EO-Bank 11

  1. Differeniate \(y=\dfrac{x}{x^2+1}\)  (2 marks)
  1. Hence evaluate \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\)   (2 marks)
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a.    \(\dfrac{dy}{dx} = \dfrac{1-x^2}{(x^2+1)^2}\)

b.    \(\dfrac{1}{2}\)

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a.    Using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{x^2+1-2x^2}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)
 

b.    Using part a: 

 \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\) \(=\left[\dfrac{x}{x^2+1}\right]_0^1\)  
  \(=\dfrac{1}{2} -0\)  
  \(=\dfrac{1}{2}\)  

 

Calculus, 2ADV C4 EO-Bank 10

Given that  `int_0^k ( 2x + 4 )\ dx = 21`, and  `k`  is a constant, find the value of  `k`.   (2 marks)

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`k = 3`

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`int_0^k ( 2x + 4 ) \ dx` `= 21`
`int_0^k ( 2x + 4 ) \ dx` `= [ x^2 + 4x ]_0^k`
  `= [(k^2 + 4k ) – 0 ]`
  `= k^2 + 4k`

 

`=> k^2 + 4k` `=21`
`k^2+4k-21` `= 0`
`(k-3)(k+7)` `= 0`
`k` `=3, -7`
`k` `=3\ text(as ) k >0`

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`
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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`