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Calculus, EXT2 C1 2023 HSC 11f

Find \({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\).  (4 marks)

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\(-\ln3 \)

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\( \text{Let}\ \ \dfrac{5x-3}{(x+1)(x-3)} = \dfrac{A}{x+1}+\dfrac{B}{x-3}\ \ \ \text{for}\ \ A, B \in \mathbb{R} \)

\(\dfrac{5x-3}{(x+1)(x-3)}\) \(=\dfrac{A}{x+1}+\dfrac{B}{x-3} \)  
  \(=\dfrac{A(x-3)+B(x+1)}{(x+1)(x-3)} \)  

 
\(\text{Equating numerators:}\)

\( A(x-3)+B(x+1)=5x-3 \)

\(\text{When}\ \ x=3, 4B=12\ \Rightarrow \ B=3 \)

\(\text{When}\ \ x=-1,  -4A=-8\ \Rightarrow \ A=2 \)

\({\displaystyle \int_0^2 \frac{5 x-3}{(x+1)(x-3)}\ dx}\) \( =\displaystyle \int_0^2 \dfrac{2}{x+1}+\dfrac{3}{x-3}\ dx \)  
  \(= \Big{[} 2\ln|x+1|+3\ln|x-3| \Big{]}_0^2 \)  
  \(= 2\ln3+3\ln1-2\ln1-3\ln3 \)  
  \(=-\ln3 \)  

Calculus, EXT2 C1 2005 HSC 1b

  1. Find real numbers  `a`  and  `b`  such that
     
    `qquad (5x)/(x^2-x-6) ≡ a/(x-3) + b/(x+2)`   (2 marks)

  2. Hence find  `int (5x)/(x^2-x-6)\ dx`   (1 mark)

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  1. `a = 3, b = 2`
  2. `3 log_e | x-3 | + 2 log_e | x+2 |+C`
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i.   `(5x)/(x^2-x-6) ≡ (a(x+2) +b(x-3))/((x-3)(x+2))`

`text(Equating numerators:)`

`5x` `=ax+2a+bx-3b`  
`5x` `=(a+b)x+2a-3b`  

 

`a+b` `=5\ \ …\ (1)`  
`2a-3b` `=0\ …\ (2)`  

 
`(1) xx 2 – (2)`

`5b` `=10`
`b` `= 2`
`a` `=3`

 

ii.    `int (5x)/(x^2-x-6)\ dx` `=int 3/(x-3)\ dx + int 2/(x+2)\ dx`
    `=3 log_e | x-3 | + 2 log_e | x+2 |+C`

Calculus, EXT2 C1 2019 HSC 11d

Find  `int 6/(x^2-9) dx`.  (3 marks)

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`ln |\ (x-3)/(x + 3)\ |+ c`

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`text(Using partial fractions):`

`6/(x^2-9)` `= 6/((x + 3)(x-3))`
  `= A/(x + 3) + B/(x-3)`

 
`:. A(x-3) + B(x + 3) = 6`
 

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

`text(When)\ \ x = -3,\ -6A = 6 \ => \ A = -1`

`int 6/(x^2-9)\ dx` `= int (-1)/(x + 3) + 1/(x-3)\ dx`
  `= -ln|\ x + 3\ | + ln |\ x-3\ | + C`
  `= ln |\ (x-3)/(x + 3)\ | + C`

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 2010 HSC 5b

Show that  `int (dy)/(y(1 − y)) = ln (y/(1 − y)) + c`

for some constant  `c`, where  `0 < y < 1`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Solution 1)`

`d/(dy)[ln(y/(1 − y)) + c]` `= d/(dy)[ln y − ln(1 − y) + c]`
  `= 1/y − (-1)/(1 − y) + 0`
  `= (1 − y + y)/(y(1 − y))`
  `= 1/(y(1 − y))`

 

`text(Solution 2)`

`text(Using partial fractions:)`

`1/(y(1 − y))=` `A/y+B/(1-y)`
   

`=> A(y-1)+By=1`

`text(When)\ \ y=1,\ B=1`

`text(When)\ \ y=0,\ A=1`

`:.int (dy)/(y(1 − y))` `= int (1/y + 1/(1 − y))\ dy`
  `= ln y − ln(1 − y) + c`
  `= ln(y/(1 − y)) + c,\ \ \ \ \ y/(1 − y) > 0\ \ text(for)\ \ 0 < y < 1.`  

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`
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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`