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Calculus, 2ADV, EQ-Bank 17

Find \(\displaystyle \int x\left(x^2+1\right)^3\, d x\).   (2 marks)

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\(\dfrac{1}{8}\left(x^2+1\right)^4+c\)

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\(\displaystyle\int x\left(x^2+1\right)^3\, d x\)

\(=\dfrac{1}{2} \displaystyle \int 2 x\left(x^2+1\right)^3 d x\)

\(=\dfrac{1}{2} \times \dfrac{1}{4}\left(x^2+1\right)^4+c\)

\(=\dfrac{1}{8}\left(x^2+1\right)^4+c\)

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. \(-2 \sqrt{x+5}+C\)
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\(B\)

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

 
\(\Rightarrow B\)

Calculus, 2ADV C4 EQ-Bank 4 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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\(C\)

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

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

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

 

\(\Rightarrow C\)

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

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

 
\( \Rightarrow A \)

Calculus, 2ADV C4 2022 HSC 18

  1. Differentiate  `y=(x^(2)+1)^(4)`.   (2 marks)

  2. Hence, or otherwise, find `int x(x^(2)+1)^(3)dx`.   (1 mark)

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a.    `dy/dx=8x(x^2+1)^3`

 b.    `1/8(x^2+1)^4+C`

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a.    `y=(x^2+1)^4`

`text{Using chain rule:}`

`dy/dx` `=4 xx 2x(x^2+1)^3`
  `=8x(x^2+1)^3`

 

b.     `intx(x^2+1)^3\ dx` `=1/8 int8x(x^2+1)^3\ dx`
    `=1/8(x^2+1)^4+C`

Calculus, 2ADV C4 EQ-Bank 5 MC

Let  `f^(prime)(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`

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`f^{prime}(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 11

Let  `f^(prime)(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`

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`f^(prime)(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`

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`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)

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a.    `5x + C`

b.    `(-3)/((x-6)) + C`

c.    `77/3`

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a.    `int 5\ dx= 5x + C`
  

b.    `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`
  

c.    `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`

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`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`

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)

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a.    `-x/sqrt(9-x^2)`

b.    `-6 sqrt(9-x^2) + C`

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IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(9-x^2)^(1/2)`.
a.     `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)`

 

b.     `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`

Calculus, 2ADV C4 2011 HSC 2e

Find  `int 1/(3x^2)\ dx`.   (2 marks)

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 `-1/(3x) + C`

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♦ Mean mark 46%.
MARKER’S COMMENT: Students who took the `1/3` out the front before integrating made less errors.
`int 1/(3x^2)\ dx` `= 1/3 int x^-2\ dx`
  `= 1/3 xx 1-1 xx x^-1 + C`
  `=-1/(3x) + C`