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Calculus, 2ADV C4 2011 HSC 4d (Adapted)

  1. Differentiate  `y=sqrt(16 -x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (8x)/sqrt(16 -x^2)\ dx`.    (2 marks)

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a.    `- x/sqrt(16-x^2)`

b.    `-8 sqrt(16-x^2) + C`

Show Worked Solution

a.    `y= sqrt(16-x^2)= (16-x^2)^(1/2)`

`dy/dx` `=1/2 xx (16-x^2)^(-1/2) xx d/dx (16-x^2)`
  `= 1/2 xx (16-x^2)^(-1/2) xx -2x`
  `= – x/sqrt(16-x^2)`

 

b.    `int (8x)/sqrt(16-x^2)\ dx` `= -8 int (-x)/sqrt(16-x^2)\ dx`
    `= -8 (sqrt(16-x^2)) + C`
    `= -8 sqrt(16-x^2) + C`

Calculus, 2ADV EO-Bank 28

  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 EO-Bank 29

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 2024 HSC 5 MC (Adapted)

What is \( {\displaystyle \int x(3 x^2+1)^4 d x} \) ?

  1. \( \dfrac{1}{30}(3x^2+1)^5+C \)
  2. \( \dfrac{1}{5}(3x^2+1)^5+C \)
  3. \( \dfrac{5}{6}(3x^2+1)^5+C \)
  4. \( \dfrac{6}{5}(3x^2+1)^5+C \)
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\( A \)

Show Worked Solution
\[ \int x(3x^2+1)^4 dx\] \(=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C\)  
  \(=\dfrac{1}{30}(3x^2+1)^5+C\)  

 
\( \Rightarrow A \)

NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.

Calculus, 2ADV EO-Bank 9 MC

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
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`=>C`

Show Worked Solution
`f^{‘}(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`