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Calculus, EXT1 C2 2024 HSC 11c

Using the substitution  \(u=x-1\), find  \(\displaystyle \int x \sqrt{x-1}\, d x\).   (3 marks)

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\(\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\)

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\(\displaystyle \int x \sqrt{x-1}\, d x \)

     
  \(u=x-1\) \(\Rightarrow \ x=u+1 \)
  \(\dfrac{d u}{d x}=1\)    \(\Rightarrow \ d u=d x\)

 

  \(\displaystyle\int(u+1) \sqrt{u+1-1}\, d u\) \(=\displaystyle{\int}(u+1) \sqrt{u} \, d u\)
    \(=\displaystyle{\int} u^{\frac{3}{2}}+u^{\frac{1}{2}}\, d u\)
    \(=\dfrac{2}{5} u^{\frac{5}{2}}+\dfrac{2}{3} u^{\frac{3}{2}}+c\)
    \(=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\)

Calculus, EXT1 C2 2022 HSC 11b

Find the exact value of  `int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx`  using the substitution `u=x^(2)+4`.  (3 marks)

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`sqrt5-2`

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`u=x^(2)+4`

`(du)/dx=2x\ \ =>\ \ du=2x\ dx`

`text{At}\ \ x=1,\ \ u=5`

`text{At}\ \ x=0,\ \ u=4`

`int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx` `=1/2 int_(4)^(5)(1)/(sqrt(u))\ du`  
  `=1/2[2sqrtu]_4^5`  
  `=[sqrtu]_4^5`  
  `=sqrt5-2`  

Calculus, EXT1 C2 2017 HSC 11e

Evaluate  `int_0^3 x/sqrt(x + 1)\ dx`, using the substitution  `x = u^2 - 1`.  (3 marks)

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`8/3`

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`x` `=u^2 – 1`
`u^2` `= x + 1`
`u` `= sqrt(x +1)`
`du` `= 1/(2sqrt(x + 1))\ dx`

 

`text(If)qquadx` `= 3,` `u` `= 2`
`x` `= 0,` `u` `= 1`
`:. int_0^3 x/sqrt(x + 1)\ dx` `= 2 int_1^2 u^2 – 1\ du`
  `= 2[(u^3)/3 – u]_1^2`
  `= 2[(8/3 – 2) – (1/3 – 1)]`
  `= 2(2/3 + 2/3)`
  `= 8/3`

Calculus, EXT1 C2 2007 HSC 1e

Use the substitution  `u = 25 - x^2`  to evaluate  `int_3^4 (2x)/(sqrt(25 - x^2))\ dx`.  (3 marks)

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`2`

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`u` `= 25 − x^2`
`(du)/(dx)` `= -2 x`
`du`  `= -2x\ dx`
`text(If)`   `x = 4,`   `u = 9`
    `x = 3,`   `u = 16`

 

`:. int_3^4 (2x)/(sqrt(25 − x^2))\ dx`

MARKER’S COMMENT: A “significant number” of students put the integral limits in the wrong order.

`= − int_16^9 u^(−1/2) du`

`= − [1/(1/2) u^(1/2)]_16^9`

`= − [2sqrtu]_16^9`

`= − [2sqrt9 − 2sqrt16]`

`= − [6 − 8]`

`= 2`

Calculus, EXT1 C2 2006 HSC 1b

Using the substitution  `u =x^4 + 8`, or otherwise, find

`int x^3 sqrt (x^4 + 8)\ dx.`  (3 marks)

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`1/6 (x^4 + 8)^(3/2) + c`

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`u = x^4 + 8`

`(du)/(dx)` `= 4x^3`
`1/4 du` `= x^3 dx` 

 
`:. int x^3 sqrt (x^4 + 8)\ dx`

`= int u^(1/2) *1/4 * du`

`= 1/4 int u^(1/2) du`

`= 1/4 * 2/3 * u^(3/2) + c`

`= 1/6 u^(3/2) + c`

`= 1/6 (x^4 + 8)^(3/2) + c`

Calculus, EXT1 C2 2005 HSC 1d

Using the substitution  `u = 2x^2 + 1`, or otherwise, find  `int x (2x^2 + 1)^(5/4)\ dx.`  (3 marks)

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`1/9 (2x^2 + 1)^(9/4) + c`

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 `u = 2x^2 + 1`

`(du)/(dx)` `= 4x`
`du` `= 4x\ dx`
`dx` `=(du)/(4x)` 

 

`:.\ int x (2x^2 + 1)^(5/4)\ dx`

`= int x * u^(5/4) * (du)/(4x)`

`= 1/4 int u^(5/4)\ du`

`= 1/4 * 4/9\ u^(9/4) + c`

`= 1/9\ u^(9/4) + c`

`= 1/9 (2x^2 + 1)^(9/4) + c`

Calculus, EXT1 C2 2014 HSC 11d

Evaluate  `int_2^5 x/(sqrt(x - 1))\ dx`  using the substitution  `x = u^2 + 1`.   (3 marks) 

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`20/3`

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`x` `= u^2 + 1`
`u^2` `= x – 1`
`u` `= sqrt (x – 1)`
`(du)/(dx)` `= 1/2 (x – 1)^(-1/2)`
  `= 1/(2 sqrt(x – 1))`
`\ \ =>2du` `= dx/sqrt(x – 1)`

 

`text(When)\ \ \ x = 5,` `\ \ u = 2`
`x = 2,` `\ \ u = 1`

`:.\ int_2^5 x/(sqrt(x – 1))\ dx`

`= 2 int_1^2 u^2 + 1\ du`

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

`= 2 [(8/3 + 2) – (1/3 + 1)]`

`= 20/3`

Calculus, EXT1 C2 2009 HSC 1f

Using the substitution  `u = x^3 + 1`, or otherwise, evaluate  `int_0^2 x^2 e^(x^3 + 1)\ dx`.   (3 marks)

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`1/3 (e^9\ – e)`

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`u` `= x^3 + 1`
`(du)/(dx)` `= 3x^2`
`du` `= 3x^2\  dx`
`text(If)\ \ \ ` `x` `= 2,\ ` `u` `= 9`
  `x` `= 0,\ ` `u` `= 1`

 

`:.\ int_0^2 x^2 e^(x^3 + 1)\ dx`

`=1/3 int_0^2 e^(x^3 + 1) * 3x^2\ dx`

`= 1/3 int_1^9 e^u\ du`

`= 1/3 [e^u]_1^9`

`= 1/3 (e^9\ – e)`