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Calculus, EXT1 C2 2023 HSC 12a

Evaluate \(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) using the substitution  \(u=x-3\).   (3 marks) 

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\(\dfrac{56}{15}\)

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\(u=x-3\ \ \Rightarrow \ x=u+3 \)

\(\dfrac{du}{dx}=1\ \ \Rightarrow \ du=dx \)

\(\text{When}\ \ x=4, u=1 \)

\(\text{When}\ \ x=3, u=0 \)

\(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) \(=\displaystyle \int_0^1(u+5) \sqrt{u}\ du\)  
  \(=\displaystyle \int_0^1 u^\frac{3}{2} +5u^\frac{1}{2}\ du\)  
  \(=\Big{[}\dfrac{2}{5} \times u^\frac{5}{2} + \dfrac{2}{3} \times 5u^\frac{3}{2}\Big{]}_0^1 \)  
  \(=\Big{[} \Big{(}\dfrac{2}{5} + \dfrac{10}{3}\Big{)}-0\Big{]}\)  
  \(=\dfrac{56}{15}\)  

Calculus, EXT1 C2 2021 HSC 11c

Use the substitution  `u = x + 1`  to find  `int xsqrt(x + 1)\ dx`.  (3 marks)

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`2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

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`u = x + 1\ \ =>\ \ (du)/(dx) = 1`

`int x sqrt(x – 1)\ dx` `= int (u – 1) sqrtu\ du`
  `= int u^(3/2) – u^(1/2)\ du`
  `= 2/5 u^(5/2) – 2/3 u^(3/2) + c`
  `= 2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

Calculus, EXT1 C2 2020 SPEC1 2

Evaluate  `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`, using the substitution  `u=1-x`.  (3 marks)

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`(8 sqrt 2)/3 – 10/3`

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`u` `= 1 – x \ => \ x = 1 – u`
`(du)/(dx)` `= -1 \ => \ dx = -du`

 

`text(When)\ \ x` `= 0,\ u = 1`
`x` `= -1,\ u = 2`

 

`int_(-1)^0 (1 + x)/sqrt(1 – x)\ dx` `= -int_2^1 (2 – u)/sqrt u\ du`
  `= int_1^2 2u^(-1/2) – u^(1/2)\ du`
  `= [4u^(1/2) – 2/3u^(3/2)]_1^2`
  `= 4 sqrt 2 – (4 sqrt 2)/3 – (4 – 2/3)`
  `= (8 sqrt 2)/3 – 10/3`

Calculus, EXT1 C2 SM-Bank 1 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
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`D`

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`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

`(du)/(dx) = 2 \ => \ dx = 1/2 du`

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>D`

Calculus, EXT1 C2 2018 HSC 11f

Evaluate  `int_-3^0 x/sqrt(1 - x) dx`, using the substitution  `u = 1 - x`.  (3 marks)

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`- 8/3`

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`u` `= 1 – x\ \ => x=1-u`
`(du)/dx` `= -1`
`dx` `= -du`

 
`text(When)\ \ x=0, \ u=1`

`text(When)\ \ x=-3, \ u=4`
 
`int_-3^0 x/sqrt (1- x)\ dx`

  `= -int_4^1 (1 – u)/sqrt(u)\ du`
  `= – int_4^1 u^(- 1/2) – u^(1/2)\ du`
  `= – [2 u^(1/2) – 2/3 u^(3/2)]_4^1`
  `= -[(2-2/3) – (2 sqrt 4 – 2/3 (sqrt 4)^3)`
  `= -[4/3 – (4-16/3)]`
  `= -8/3`

Calculus, EXT1 C2 2016 HSC 11b

Use the substitution  `u = x - 4`  to find  `int xsqrt(x - 4)\ dx`.  (3 marks)

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`2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

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`u = x – 4\ \ => \ x = u + 4`

`(du)/(dx) = 1\ \ => \ dx = du`
 

`:. int x sqrt (x – 4)\ dx`

`= int (u + 4) · u^(1/2)\ du`

`= int u^(3/2) + 4u^(1/2)\ du`

`= 2/5 u^(5/2) + 4 · 2/3 u^(3/2) + c`

`= 2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

Calculus, EXT1 C2 2004 HSC 1e

Use the substitution  `u = x − 3` to evaluate

`int_3^4 xsqrt(x − 3)\ dx.` (3 marks)

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`2 2/5`

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`text(Let)\ \ u = x − 3`

`=> x = u + 3`

`(du)/dx = 1`

`=> dx = du`

`text(When)\ \ ` `x = 4,` `u = 1`
  `x = 3,` `u = 0`

 
`int_3^4 xsqrt(x − 3\ dx)`

`= int_0^1(u + 3)\ u^(1/2)\ du`

`= int_0^1u^(3/2) + 3u^(1/2)\ du`

`=[2/5u^(5/2) + 3 xx 2/3u^(3/2)]_0^1`

`= [2/5u^(5/2) + 2u^(3/2)]_0^1`

`= [(2/5 + 2) − 0]`

`= 2 2/5`

Calculus, EXT1 C2 2015 HSC 11e

Use the substitution  `u = 2x - 1`  to evaluate  `int_1^2 x/((2x - 1)^2)\ dx`.  (3 marks)

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`1/4(ln 3 + 2/3)`

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`u = 2x − 1`

`⇒ 2x` `= u + 1`
 `x` `= 1/2(u + 1)`
`(du)/(dx)` `= 2`
`dx` `= (du)/2`
`text(When)` `\ \ x = 2,\ ` `u = 3`
  `\ \ x = 1,\ ` `u = 1`

 

`:. int_1^2 x/((2x − 1)^2) \ dx`

`= int_1^3 1/2(u + 1) · 1/(u^2) · (du)/2`

`= 1/4int_1^3 ((u + 1)/(u^2)) du`

`= 1/4 int_1^3 1/u + u^(−2) du`

`= 1/4 [ln u − u^(−1)]_1^3`

`= 1/4 [(ln 3 − 1/3) − (ln 1 − 1)]`

`= 1/4 (ln 3 − 1/3 + 1)`

`= 1/4(ln 3 + 2/3)`

Calculus, EXT1 C2 2013 HSC 5 MC

Which integral is obtained when the substitution  `u = 1 + 2x`  is applied to  `int x sqrt(1 + 2x)\ dx`?

  1. `1/4 int (u - 1) sqrt u\ du`
  2. `1/2 int (u - 1) sqrt u\ du` 
  3. `int (u - 1) sqrt u\ du`
  4. `2 int (u - 1) sqrt u\ du`
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`A`

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`text(Let)\ \ u` `= 1 + 2x`
`:.x` `= 1/2 (u – 1)`
`(du)/(dx)` `= 2`
`:.dx` `= 1/2\ du`

 
`int x sqrt(1 + 2x)\ dx`

`=int 1/2 (u – 1) xx u^(1/2) xx 1/2\ du`

`= 1/4 int (u – 1) sqrt u\ du`
 

`=>  A`

Calculus, EXT1 C2 2010 HSC 1e

Use the substitution  `u = 1 - x`  to evaluate  `int_0^1 x sqrt(1 - x)\ dx`.   (3 marks)

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`4/15`

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`u = 1 – x` `\ \ \ \ \ => x = 1 – u`
`(du)/(dx) = -1` `\ \ \ \ \ => du = – dx`

 

`text(When)\ \ \ \ ` `x = 1,\ \ ` `u = 0`
  `x = 0,\ \ ` `u = 1`

 
`:. int_0^1 x sqrt(1 – x)\ dx`

`= – int_1^0 (1 – u) u^(1/2)\ du`

`= int_1^0 (u – 1) u^(1/2)\ du`

`= int_1^0 (u^(3/2) – u^(1/2))\ du`

`= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^0`

`= [0 – (2/5 – 2/3)]`

`= – (6/15 – 10/15)`

`= 4/15`

Calculus, EXT1 C2 2012 HSC 11d

Use the substitution  `u = 2 - x`  to evaluate  `int_1^2 x (2 - x)^5\ dx`.   (3 marks)  

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 `4/21`

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`u` `=2-x`
`:. x` `= 2 – u`
`(du)/dx` `= -1`
`:.dx` `=-du`

 

`text(When)\ \ x = 2,` `\ \ u = 0`
`text(When)\ \ x = 1,` `\ \ u = 1`

 

`:. int_1^2 x (2 – x)^5\ dx` `= int_1^0 – (2 – u) u^5\ du`
  `= int_1^0 u^6 – 2u^5\ du`
  `= [1/7 u^7 – 2/6 u^6]_1^0`
  `= [0 – (1/7 – 1/3)]`
  `= – (3/21 – 7/21)`
  `= 4/21`