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Calculus, EXT1 C2 EQ-Bank 3

Use the substitution  \(u=\cos\,x\)  to evaluate

\(\displaystyle {\int}_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin x\, dx\).   (3 marks)

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\(1-\dfrac{1}{e}\)

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\(u=\cos x\)

\(\dfrac{d u}{d x}=-\sin\,x\  \Rightarrow\  du=-\sin\,x\,dx\)

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

\(\text{When} \ \ x=\dfrac{\pi}{2}, u=0\)

\(\displaystyle \int_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin\,x\,d x\) \(=\displaystyle -\int_0^{-1} e^u\,d u\)
  \(=\left[-e^u\right]_0^{-1}\)
  \(=-e^{-1}+e^0\)
  \(=1-\dfrac{1}{e}\)

Calculus, EXT1 C2 2019 HSC 13a

Use the substitution  `u = cos^2 x`  to evaluate  `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`.  (3 marks)

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`ln {:10/9`

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`u` `= cos^2 x`
`(du)/(dx)` `= -2 sin x cos x`
  `= -sin 2x`
`du` `= -sin 2x\ dx`

 
`text(When)\ \ x = pi/4,\ \ u = 1/2`

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

`:. int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx` `= -int_1^(1/2) (du)/(4 + u)`
  `= -[ln (4 + u)]_1^(1/2)`
  `= -(ln 4.5 – ln 5)`
  `= -ln {:9/10`
  `= ln {:10/9`

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 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 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`