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Calculus, 2ADV C4 2011 HSC 4d v1

  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)

Show Answers Only
  1. `- x/sqrt(16\ -x^2)`
  2. `-8 sqrt(16\ -x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(16-x^2)^(1/2)`.
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 C4 2024 HSC 5 MC v1

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 C4 2025 HSC 5 MC

What is  \(\displaystyle \int \frac{1}{\sqrt{x+5}} d x\) ?

  1. \(\dfrac{1}{2} \sqrt{x+5}+C\)
  2. \(2 \sqrt{x+5}+C\)
  3. \(-\dfrac{1}{2} \sqrt{x+5}+C\)
  4. \(-2 \sqrt{x+5}+C\)
Show Answers Only

\(B\)

Show Worked Solution
\(\displaystyle \int \frac{1}{\sqrt{x+5}}\ d x\) \(=\displaystyle \int (x+5)^{-\frac{1}{2}}\ d x\)  
  \(= 2 \sqrt{x+5}+C\)  

 
\(\Rightarrow B\)

Calculus, 2ADV C4 EO-Bank 4 MC SJ

 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`
Show Answers Only

`=>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`

Calculus, 2ADV C4 EQ-Bank 1 MC

What is \(\displaystyle \int x\left(4x^2+2\right)^3 dx\)

  1. \(8x\left(4 x^2+2\right)^2+c\)
  2. \(\dfrac{1}{12} x\left(4 x^2+2\right)^4+c\)
  3. \(\dfrac{1}{32}\left(4 x^2+2\right)^4+c\)
  4. \(\dfrac{1}{4} x\left(4 x^2+2\right)^4+c\)
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\(\Rightarrow C\)

Show Worked Solution

\(\text{Strategy: differentiate answers}\)

\(\text {Option C} \ \ \Rightarrow \ \ \text{using product and chain rules:}\)

\(\genfrac{}{}{0pt}{}{\Large{\ \ d}}{\Large{dx}}\left(\dfrac{1}{32}\left(4 x^2+2\right)^4\right)\) \(=\dfrac{1}{32} \times 4 \times 8 x\left(4 x^2+2\right)^3\)
  \(=x\left(4 x^2+2\right)^3\)

 

\(\Rightarrow C\)

Calculus, 2ADV C4 2024 HSC 5 MC

What is \( {\displaystyle \int(6 x+1)^3 d x} \) ?

  1. \( \dfrac{1}{24}(6 x+1)^4+C \)
  2. \( \dfrac{1}{4}(6 x+1)^4+C \)
  3. \( \dfrac{2}{3}(6 x+1)^4+C \)
  4. \( \dfrac{3}{2}(6 x+1)^4+C \)
Show Answers Only

\( A \)

Show Worked Solution
\[ \int(6 x+1)^3 dx\] \(=\dfrac{1}{4} \cdot \dfrac{1}{6}(6 x+1)^4+C\)  
  \(=\dfrac{1}{24}(6 x+1)^4+C\)  

 
\( \Rightarrow A \)

Calculus, 2ADV C4 SM-Bank 4 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`
Show Answers Only

`=>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`

Calculus, 2ADV C4 SM-Bank 5

Let  `f^(′)(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

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`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f^(′)(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Calculus, 2ADV C4 2019 MET1 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

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`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

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)

Show Answers Only
  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 2011 HSC 4d

  1. Differentiate  `y=sqrt(9 - x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (6x)/sqrt(9 - x^2)\ dx`.    (2 marks)

Show Answers Only
  1. `- x/sqrt(9\ – x^2)`
  2. `-6 sqrt(9\ – x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(9-x^2)^(1/2)`.
i.    `y` `= sqrt(9 – x^2)`
    `= (9 – x^2)^(1/2)`

 

`dy/dx` `=1/2 xx (9 – x^2)^(-1/2) xx d/dx (9 – x^2)`
  `= 1/2 xx (9 – x^2)^(-1/2) xx -2x`
  `= – x/sqrt(9 – x^2)`

 

ii.    `int (6x)/sqrt(9 – x^2)\ dx` `= -6 int (-x)/sqrt(9 – x^2)\ dx`
    `= -6 (sqrt(9 – x^2)) + C`
    `= -6 sqrt(9 – x^2) + C`