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Calculus, EXT2 C1 2008 HSC 1e

It can be shown that

`qquad (8(1-x))/((2-x^2)(2-2x+x^2)) = (4-2x)/(2-2x+x^2)-(2x)/(2-x^2)`    (do NOT prove this)
 

Use this result to evaluate  `int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`   (4 marks)

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`pi/2`

Show Worked Solution

`int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`

`=int_0^1 (4-2x)/(2-2x+x^2)\ dx-int_0^1 (2x)/(2-x^2)\ dx`

`=int_0^1 (2-(2x-2))/(2-2x+x^2)\ dx + [log_e|2-x^2|]_0^1`

`=int_0^1 2/(1+(1-x)^2)-int_0^1 (2x-2)/(2-2x+x^2) + [log_e|2-x^2|]_0^1`

`=[-2tan^(-1)(1-x)]_0^1-[log_e(2-2x+x^2)]_0^1 + [log_e|2-x^2|]_0^1`

`=-2tan^(-1)0+2tan^(-1) 1-(log_e1-log_e2) + (log_e 1-log_e2)`

`=0+2 xx pi/4 + log_e2-log_e 2`

`=pi/2`

Calculus, EXT2 C1 2003 HSC 1d

  1.  Find the real numbers  `a`  and  `b`  such that
     
    `qquad (5x^2-3x+13)/((x-1)(x^2+4)) ≡ a/(x-1) + (bx-1)/(x^2+4)`   (2 marks)

  2.  Hence find  `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx`   (2 marks)

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  1. `a = 3, b = 2`
  2. `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`
Show Worked Solution

i.   `(5x^2-3x+13)/((x-1)(x^2+4)) ≡ (a(x^2+4) + (bx-1)(x-1))/((x-1)(x^2+4))`

 
`text(Equating numerators:)`

`5x^2-3x+13` `=ax^2+4a+bx^2-bx-x+1`  
  `=(a+b)x^2+(-b-1)x+4a+1`  

 

`-b-1` `=-3\ \ =>\ \ b=2`  
`a` `=3`  

 

ii.    `int (5x^2-3x+13)/((x-1)(x^2+4)) \ dx` `=int 3/(x-1)\ dx-int (2x)/(x^2+4)\ dx-int 1/(4+x^2)\ dx`
    `=3 log_e|x-1|-log_e |x^2+4|-1/2 tan^(-1)(x/2)+C`

Calculus, EXT2 C1 2017 HSC 14a

It is given that  `x^4 + 4 = (x^2 + 2x + 2) (x^2-2x + 2)`.

  1. Find `A` and `B` so that  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`.  (1 mark)

  2. Hence, or otherwise, show that for any real number `m`
     
         `int_0^m 16/(x^4 + 4)\ dx = ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`.  (2 marks)

  3. Find the limiting value as  `m -> oo`  of
     
         `int_0^m 16/(x^4 + 4)\ dx`.  (1 mark)

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  1. `A = 4 and B = 4`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2 pi`
Show Worked Solution

i.  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`

`text(Multiply each side by)\ \ x^4 + 4`

`16 = (A + 2x) (x^2-2x + 2) + (B-2x) (x^2 + 2x + 2)`

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

`2A + 2B` `=16`
`A + B` `= 8\ text{… (1)}`

 
`text(When)\ \ x = 1`

`(A + 2) (1) + (B-2) (5)` `=16`
`A + 2 + 5B-10` `=16`
`A + 5B` `= 24\ text{… (2)}`

 

`text{Subtract  (2) – (1)}`

`4B=16\ \ =>\ \ B=4`

`A=4`
 

ii.  `int_0^m 16/(x^4 + 4)`

`= int_0^m (4 + 2x)/(x^2 + 2x + 2) dx + int_0^m (4-2x)/(x^2-2x + 2) dx`

`= int_0^m (2x + 2)/(x^2 + 2x + 2) + 2/(x^2 + 2x + 2) dx + int_0^m 2/(x^2-2x + 2)-(2x-2)/(x^2-2x + 2) dx`

`= [ln(x^2 + 2x + 2)]_0^m + int_0^m 2/(1 + (x + 1)^2) dx + int_0^m 2/(1 + (x-1)^2) dx-[ln (x^2-2x + 2)]_0^m`

`= [ln(m^2 + 2m + 2)-ln 2] + [2 tan^(-1) (x + 1)]_0^m + [2 tan^(-1) (x-1)]_0^m-[ln (m^2-2m + 2)-ln 2]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 [tan^(-1) (m + 1)-tan^(-1) (1)] + 2 [tan^(-1) (m-1)-tan^(-1) (1)]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1)-2 · pi/4 + 2 tan ^(-1) (m-1) + 2 · pi/4`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

 

iii.  `I = int_0^m 16/(x^4 + 4) = ln ((1 + 2/m + 2/m^2)/(1-2/m + 2/m^2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

`lim_(m -> oo) I` `= ln\ 1 + 2 · pi/2 + 2 · pi/2`
  `= 0 + pi + pi`
  `= 2 pi`