SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, EXT2 C1 EQ-Bank 20

Find  \(\displaystyle \int_0^1\dfrac{3}{2 \log _e 2}\left(\dfrac{1}{(t+1)(2-t)}\right) d t\)   (4 marks)

Show Answers Only

\(I=\dfrac{1}{2}\)

Show Worked Solution

\(I=\dfrac{3}{2 \ln 2} \displaystyle \int_0^1 \dfrac{t}{(t+1)(2-t)}\, dt\)
 

\(\text {Using partial fractions:}\)

\(\dfrac{t}{(t+1)(2-t)}=\dfrac{A}{(t+1)}+\dfrac{B}{(2-t)}\)

\(A(2-t)+B(t+1)=t\)

\(\text{If}\ \ t=2:\)

\(3 B=2 \ \Rightarrow \ B=\dfrac{2}{3}\)

\(\text{If}\ \ t=-1:\)

\(3 A=-1 \ \Rightarrow \ A=-\dfrac{1}{3}\)
 

\(I\) \(=\displaystyle \frac{3}{2 \ln 2} \int_0^1-\frac{1}{3(t+1)}+\frac{2}{3(2-t)} d t\)
  \(=\displaystyle\frac{1}{2 \ln 2} \int_0^1-\frac{1}{t+1}+\frac{2}{2-t} d t\)
  \(=\displaystyle\frac{1}{2 \ln 2}\Big[-\ln \abs{t+1}+2 \ln \abs{2-t}\Big]_0^1\)
  \(=\displaystyle\frac{1}{2 \ln 2}[(-\ln 2+2 \ln 1)-(2 \ln 1-2 \ln 2)]\)
  \(=\displaystyle\frac{1}{2 \ln 2}(\ln 2)\)
  \(=\displaystyle\frac{1}{2}\)

Calculus, EXT2 C1 2025 HSC 12d

Find \(\displaystyle \int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)} \, dx\).   (4 marks)

Show Answers Only

\(2\, \tan ^{-1} x-\ln \abs{4-x}+C\)

Show Worked Solution

\(\dfrac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}=\dfrac{A}{(4-x)}+\dfrac{B x+C}{x^2+1}\)

\(A\left(x^2+1\right)+(B x+C)(4-x)=x^2-2 x+9\)

\(\text{If}\ \  x=4:\)

\(17 A=16-8+9=17 \ \ \Rightarrow\ \ A=1\)

\(\text{If}\ \  x=0:\)

\(1+c \times 4=9 \ \ \Rightarrow\ \ C=2\)

\(\text{If}\ \  x=1:\)

\(2+3 B+6=8 \ \ \Rightarrow\ \ B=0\)

\(\displaystyle\int \frac{x^2-2 x+9}{(4-x)\left(x^2+1\right)}\, d x\) \(=\displaystyle\int \frac{1}{4-x}\, d x+\int \frac{2}{x^2+1}\, d x\)
  \(=2\, \tan ^{-1} x-\ln \abs{4-x}+C\)

Calculus, EXT2 C1 2024 HSC 12b

Use partial fractions to find \(\displaystyle \int \frac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x\)   (3 marks)

Show Answers Only

\(I=3 \ln \abs{x-1}+2 \tan ^{-1}(x)+c\)

Show Worked Solution

\(\displaystyle\int \dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}\, d x\)

\(\dfrac{3 x^2+2 x+1}{(x-1)\left(x^2+1\right)}=\dfrac{A}{(x-1)}+\dfrac{B x+C}{x^2+1}\)

  \(3 x^2+2 x+1\) \(=A\left(x^2+1\right)+(x-1)(B x+C)\)
    \(=A x^2+A+B x^2+C x-B x-C\)
    \(=(A+B) x^2+(C-B) x+A-C\)

 
\(\text {If } x=1:\)

\(3+2+1=2 A \ \Rightarrow \ A=3\)

\(A+B=3 \quad \quad \ \Rightarrow \ B=0\)

\(A-C=1 \quad \quad \ \Rightarrow \ C=2\)

  \(\therefore I\) \(=\displaystyle \int \frac{3}{x-1}+\frac{2}{x^2+1}\, d x\)
    \(=3 \ln \abs{x-1}+2 \tan ^{-1}(x)+c\)

Calculus, EXT2 C1 2022 SPEC1 4

Find `int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx`.   (4 marks)

Show Answers Only

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

Show Worked Solution

`text{Using partial fractions:}`

`(3x^(2)+4x+12)/(x(x^(2)+4))` `-=(A)/(x)+(Bx+C)/(x^(2)+4)`  
`3x^(2)+4x+12` `=A(x^(2)+4)+x(Bx+C)`  
`3x^(2)+4x+12` `=(A+B)x^(2)+Cx+4A`  

 
`4A=12\ \ =>\ \ A=3`

`C=4`

`A+B=3\ \ =>\ \ B=0`

`:.\int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx` `=int(3)/(x)+(4)/(4+x^(2))\ dx`  
  `=3ln |x|+2tan^(-1)((x)/(2))+c`  
Mean mark 55%.

Calculus, EXT2 C1 2023 HSC 11f

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

Show Answers Only

\(-\ln3 \)

Show Worked Solution

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

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

 
\(\text{Equating numerators:}\)

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

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

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

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

Calculus, EXT2 C1 2022 HSC 12d

Using partial fractions, evaluate  `int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`, giving your answer in the form  `(1)/(2)ln((f(n))/(8(n-1)^(2)))`, where `f(n)` is a function of `n`.  (4 marks)

Show Answers Only

`1/2ln((4+n^2)/(8(1-n^2)))`

Show Worked Solution
`(4+x)/((1-x)(4+x^(2)))` `≡ A/(1-x) + (Bx+C)/(4+x^2)`  
`4+x` `≡A(4+x^2)+(Bx+C)(1-x)`  

 
`text{If}\ \ x=1, \ 5=5A\ \ =>\ \ A=1`

`(4+x)` `≡ 4+x^2+Bx-Bx^2+C-Cx`  
`4+x` `≡ (1-B)x^2+(B-C)x+C+4`  

 
`=>\ B=1, \ C=0`
 


Mean mark 85%.

`:.int_(2)^(n)(4+x)/((1-x)(4+x^(2))) dx`

`=int_2^n 1/(1-x) +x/(4+x^2)\ dx`

`=[-ln abs(1-x)+1/2ln(4+x^2)]_2^n`

`=-ln abs(1-n)+1/2ln(4+n^2)+lnabs(1-2)-1/2ln(4+2^2)`

`=-1/2ln(1-n)^2+1/2ln(4+n^2)-1/2ln(8)`

`=1/2ln((4+n^2)/(8(1-n^2)))`

Calculus, EXT2 C1 2021 HSC 11f

Express  `{3x^2-5}/{(x-2)(x^2 + x + 1)}`  as a sum of partial fractions over `RR`.  (3 marks)

Show Answers Only

`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

Show Worked Solution

`{3x^2-5}/{(x-2)(x^2 + x + 1)} = {A}/{(x-2)} + {B x + C}/{(x^2 + x + 1)}`

`A (x^2 + x + 1) + (Bx + C)(x-2) ≡ 3x^2-5`

`text{If} \ \ x = 2,`

`7A = 7 \ => \ A = 1`

`text{If} \ \ x = 0,`

`1-2C=-5 \ => \ C = 3`

`text(Equating coefficients:)`

`x^2 + x + 1 + Bx^2-2Bx + 3x -6 \ ≡ \ 3x^2 -5`

`(B + 1) x^2 + (4 -2B) x -5 \ ≡ \ 3x^2-5`

`=> B = 2`
 

`:. \ {3x^2-5}/{(x-2)(x^2 + x + 1)} = {1}/{x-2} + {2x + 3}/{x^2 + x + 1}`

Calculus, EXT2 C1 EQ-Bank 18

Using partial fractions, show that

`int_0^(1/2) 8/(1-x^4)\ dx = log_e 9-4 tan^(−1) (1/2)`  (3 marks)

Show Answers Only

`text(See Worked Solutions)`

Show Worked Solution

`text(Using partial fractions:)`

`8/(1-x^4) = A/(1-x^2) + B/(1 + x^2)`

`A(1 + x^2) + B(1-x^2)` `= 8`
`A + B + (A-B)x^2` `= 8`
`A + B` ` = 8\ \ …\ (1)`
`A-B` ` = 0\ \ …\ (2)`

 
`A = 4, \ B = 4`

 

`4/(1-x^2) = A/(1-x) + B/(1 + x)`

`A(1 + x) + B(1-x)` `= 4`
`A + B + (A-B)x` `= 4`
`A + B` `= 4\ \ …\ (1)`
`A-B` `= 0\ \ …\ (2)`

 
`A = 2, B = 2`
 

`int_0^(1/2) 8/(1-x^4)\ dx` `= int_0^(1/2) 2/(1-x) + 2/(1 + x) + 4/(1 + x^2)\ dx`
  `= [−2ln |1-x| + 2ln |1 + x| + 4tan^(−1)x]_0^(1/2)`
  `= [2ln |(1 + x)/(1-x)| + 4tan^(−1)x]_0^(1/2)`
  `= 2ln |(1 1/2)/(1/2)| + 4tan^(−1)(1/2)-(2ln1 + 4tan^(−1) 0)`
  `= 2ln3 + 4tan^(−1)(1/2)`
  `= ln9 + 4tan^(−1)(1/2)`

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)

Show Answers Only

`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)

Show Answers Only

a.    `a = 3, b = 2`

b.    `3 log_e|x-1| + log_e |x^2+4|-1/2 tan^(-1)(x/2) +C`

Show Worked Solution

a.   `(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`  

 

b.     `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 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)

Show Answers Only

a.    `a = 3, b = 2`

b.    `3 log_e | x-3 | + 2 log_e | x+2 |+C`

Show Worked Solution

a.    `(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`

 

b.     `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)

Show Answers Only

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

Show Worked Solution

`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 2018 HSC 11c

By writing  `(x^2 - x - 6)/((x + 1)(x^2 - 3))`  in the form  `a/(x + 1) + (bx + c)/(x^2 - 3)`,

find  `int(x^2 - x - 6)/((x + 1)(x^2 - 3))\ dx`.  (4 marks)

Show Answers Only

`2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + c`

Show Worked Solution
`(x^2 – x – 6)/((x + 1)(x^2 – 3))` `≡ a/(x + 1) + (bx + c)/(x^2 – 3)`
`x^2 – x – 6` `≡ a(x^2 – 3) + (bx + c)(x + 1)`

 
`text(When)\ x = −1`

`1 + 1 – 6 = −2a\ \ =>\ \ a = 2`
 

`text(Equating co-efficients of)\ x^2`

`1 = (a + b)x^2\ \ =>\ \ b = −1`
 

`text(Equating co-efficients of)\ x`

`−1 = b + c\ \ =>\ \ c = 0`
 

`:. int(x^2 – x – 6)/((x + 1)(x^2 – 3))\ dx` `= int2/(x + 1) – x/(x^2 – 3)\ dx`
  `= 2 ln |\ x + 1\ | – 1/2 ln |\ x^2 – 3\ | + 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)

Show Answers Only
  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`

Calculus, EXT2 C1 2007 HSC 1e

It can be shown that

`2/(x^3 + x^2 + x + 1) = 1/(x + 1) - x/(x^2 + 1) + 1/(x^2 + 1).`   (Do NOT prove this.)
 

Use this result to evaluate  `int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx.`  (4 marks)

 

Show Answers Only

`tan^-1 2 – tan^-1­ 1/2`

Show Worked Solution

`int_(1/2)^2 2/(x^3 + x^2 + x + 1)\ dx`

`=int_(1/2)^2 (1/(x + 1) – x/(x^2 + 1) + 1/(x^2 + 1)) dx`

`=[log_e(x + 1) – 1/2 log_e (x^2 + 1) + tan^-1 x]_(1/2)^2`

`=[(log_e 3 – 1/2 log_e 5 + tan^-1 2) – (log_e­ 3/2 – 1/2 log_e­ 5/4 + tan^-1­ 1/2)]`

`=log_e\ 3/sqrt5 -log_e (3/2 xx 2/sqrt5) + tan^-1 2 – tan^-1­ 1/2`

`=log_e (3/sqrt(5) xx sqrt (5)/3) + tan^-1 2 – tan^-1­ 1/2`

`=tan^-1 2 – tan^-1­ 1/2`

Calculus, EXT2 C1 2006 HSC 1c

  1. Given that  `(16x - 43)/((x - 3)^2 (x + 2))`  can be written as
     
    `qquad (16x - 43)/((x - 3)^2 (x + 2)) = a/(x - 3)^2 + b/(x - 3) + c/(x + 2)`,
     
    where  `a, b` and `c`  are real numbers, find  `a, b and c.`  (3 marks)

  2. Hence find  `int (16x - 43)/((x - 3)^2 (x + 2))\ dx.`  (2 marks)

Show Answers Only

a.    `a = 1, b = 3, c = -3`

b.    `-1/(x-3)+3ln((x-3)/(x+2)) +c`

Show Worked Solution

a.    `(16x – 43)/((x – 3)^2 (x + 2)) = a/(x – 3)^2 + b/(x – 3) + c/(x + 2)`

`16x – 43 = a (x + 2) + b (x – 3) (x + 2) + c (x – 3)^2`
 

`text(When)\ \ x = 3,\ \ 5a =5\ \ =>a=1`

`text(When)\ \ x=–2,\ \ 25c=–75\ \ =>c=–3`

`text(When)\ \ x=0`

`-43` `= 2(1) – 6b + (-3)(-3)^2`
`6b` `= 18`
`b` `=3`

 
`:.a=1, b=3, c=–3`
 

b.    `int (16x – 43)/((x – 3)^2 (x + 2))\ dx`

`=int (1/(x – 3)^2 + 3/(x – 3) – 3/(x + 2))\ dx`

`=-1/(x – 3) + 3ln(x – 3) -3ln(x + 2) + c`

`=-1/(x-3)+3ln((x-3)/(x+2)) +c`

Calculus, EXT2 C1 2009 HSC 1d

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

Show Answers Only

`2 ln (3/4)`

Show Worked Solution

`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)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution

`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 2010 HSC 1c

Find  `int 1/(x(x^2 + 1))\ dx`.   (3 marks)

Show Answers Only

`ln\ |\ x\ | + 1/2ln\ (x^2 + 1) + c`

Show Worked Solution

`text(Using partial fractions:)`

`1/(x(x^2 + 1)) =` `a/x + (bx + c)/(x^2 + 1)`
`1=` `a(x^2+1)+x(bx+c)`
`1=` `(a + b)x^2 + cx + a`
`:.c = 0, \ \ a = 1,\ \ b = -1`

 

`:.int 1/(x(x^2 + 1))\ dx` `=int(1/x − x/(x^2 + 1))\ dx`
  `=ln\ |\ x\ |-1/2ln\ (x^2 + 1) + c`

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)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.     `1/4 log_e 3`

Show Worked Solution

a.    `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`

 

b.    `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`

Calculus, EXT2 C1 2011 HSC 1c

  1. Find real numbers `a, b` and `c` such that 
      
       `1/(x^2 (x - 1)) = a/x + b/x^2 + c/(x - 1).`  (2 marks)

  2. Hence, find  `int 1/(x^2 (x - 1))\ dx`  (2 marks)

Show Answers Only

a.    `a = −1,\ \ \ b = −1,\ \ \ c = 1`

b.    `log_e\ ((x – 1)/x) + 1/x + c`

Show Worked Solution

a.    `1/(x^2 (x – 1)) = a/x + b/x^2 + c/(x – 1)`

`1 = ax (x – 1) + b (x – 1) + cx^2`

`1 = ax^2 + cx^2 – ax + bx – b`

`1=(a+c)x^2+(b-a)x-b`
 

`text(Equating coefficients:)`

`=>0 = a + c,\ \ \ b – a = 0,\ \ \ -b = 1`

`:.a = -1,\ \ \ b = -1,\ \ \ c = 1`

 

b.   `int 1/(x^2 (x – 1)) \ dx` `= int(-1/x – 1/x^2 + 1/(x – 1))\ dx`
  `= -log_e x + 1/x + log_e (x – 1) + c`
  `= log_e\ ((x – 1)/x) + 1/x + c`