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Calculus, SPEC1 2011 VCAA 1

Find an antiderivative of  `(1 + x)/(9-x^2),\ \ x in R \ text(\{– 3, 3}).`  (3 marks)

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`-1/3ln((3-x)^2(|3 + x|))`

Show Worked Solution
`int (1 + x)/(9-x^2)\ dx` `= int (1-x)/((3-x)(3 + x))\ dx`
  `= int A/(3-x) + B/(3 + x)\ dx`

 
`text(Using partial fractions:)`

`A(3 + x) + B(3-x) = 1 + x`

`text(When)\ \ x=3, \ 6A=4\ \ =>\ \ A=2/3`

`text(When)\ \ x=-3, \ 6B=-2\ \ =>\ \ B=-1/3`
 

`:. int (1 + x)/(9-x^2)\ dx`

`int 2/3 (1/(3-x))-1/3 (1/(3 + x))\ dx`

`= -2/3ln|3-x|-1/3ln|3 + x|,\ \ \ (c=0)`

Calculus, SPEC1 2013 VCAA 2

Evaluate  `int_0^1 (x-5)/(x^2-5x + 6)\ dx.`  (4 marks)

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`ln(9/32)`

Show Worked Solution
`int_0^1 (x-5)/(x^2-5x + 6)\ dx` `= int_0^1 (x-5)/((x-3)(x-2))\ dx`
  `= int_0^1 A/(x-3) + B/(x-2)\ dx`

 
`text(Using partial fractions:)` 

`A(x-2) + B(x-3) = x-5`

`text(If)\ \ x = 2,` `-B` `= −3`
  `B` `= 3`
`text(If)\ \ x = 3,` `A` `=-2`

 
`:. int_0^1 (x-5)/(x^2-5x + 6)\ dx`

  `= int_0^1 −2/(x-3) + 3/(x-2)\ dx`
  `= [−2ln|x-3| + 3ln|x-2|]_0^1`
  `= −2ln(2) + 3ln(1) + 2ln(3)-3ln(2)`
  `= 2ln(3)-5ln(2)`
  `= ln(3^2/2^5)`
  `= ln(9/32)`

Calculus, SPEC1-NHT 2017 VCAA 2

Find  `a`  given that  `int_(-2)^a 8/(16-x^2)\ dx = log_e(6), \ a in (-2, 4)`.  (3 marks)

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 `a = 4/3`

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`I` `= int_(-2)^a 8/((4-x)(4 + x))\ dx`
  `= int_(-2)^a A/(4-x) + B/(4 + x)\ dx`

 
`text(Using partial fractions:)`

`A(4 + x) + B(4-x) = 8`

`text(If)\ \ x = 4\ \ =>\ \ 8A=8\ \ =>\ \ A=1`

`text(If)\ \ x = -4\ \ =>\ \ 8B=8\ \ =>\ \ B=1`
 

`:. I` `= int_(-2)^a 1/(4-x) + 1/(4 + x)\ dx`
`ln6` `= [ln |4 + x|-ln|4-x|]_(-2)^a`
  `= ln|4 + a|-ln|4-a|-(ln|2|-ln|6|)`
  `= ln ((4 + a)/(4-a))-ln (2/6),\ \ \ (a in (-2, 4))`
  `= ln ((3(4 + a))/(4-a))`
`6` `= 3((4 + a)/(4-a))`
`2` `=(4 + a)/(4-a)`
`8-2a` `=4+a`
`:. a` `=4/3`

Calculus, EXT2 C1 2009 HSC 1d

Evaluate  `int_2^5 (x-6)/(x^2 + 3x-4)\ dx.`   (4 marks)

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`2 ln (3/4)`

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`text(Using partial fractions:)`

`(x-6)/(x^2 + 3x-4)` `=(x-6)/((x+4)(x-1))`
  `= A/(x+4) + B/(x-1)`
`:. x-6` `=A(x-1)+B(x+4)`

 

`text(When)\ \ x=1,\ \ 5B=-5\ =>B=-1`

`text(When)\ \ x=-4,\ \ -5A=-10\ =>A=2`

`:. int_2^5 (x-6)/(x^2 + 3x-4)\ dx` `= int_2^5 2/(x+4)\ dx-int_2^5 (dx)/(x-1)`
  `= [2 ln (x+4)]_2^5 -[ln(x-1)]_2^5` 
  `= 2(ln 9-ln 6)-(ln4-ln1)`
  `= 2ln(3/2)-ln4`
  `= 2ln(3/2)-2ln2`
  `= 2ln(3/4)`

Calculus, EXT2 C1 2011 HSC 7b

Let   `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

  1. Use the substitution  `u = 4-x`  to show that
     
          `I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.`  (2 marks)

     

  2. Hence, find the value of  `I`.  (3 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/4 log_e 3`
Show Worked Solution

i.   `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

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

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

`I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
  `=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
  `=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
  `=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

 

ii.   `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

 

`text(We can add the integrals such that)`

`2I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
  `=int_1^3 1/(x(4-x)) dx`

 

`text(Using partial fractions:)`

♦ Mean mark 36%.
`1/(x(4-x))` `=A/x+B/(4-x)`
`1` `=A(4-x)+Bx`

 
`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I` `=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
  `=1/4 [log_e x-log_e (4-x)]_1^3`
  `=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
  `=1/2 log_e 3`
`:.I` `=1/4 log_e 3`