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Calculus, EXT2 C1 2019 HSC 15c

  1. Show that  `int_0^1 x/(x + 1)^2\ dx = ln 2 - 1/2`.  (2 marks)


  2. Let  `I_n = int_0^1 x^n/(x + 1)^2\ dx`.

     


    Show that  `I_n = 1/(2(n - 1)) - n/(n - 1) I_(n - 1)\ \ text(for)\ \ n >= 2`.  (3 marks)


  3. Evaluate  `I_3`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.    `text(Show)\ \ int_0^1 x/(x + 1)^2\ dx = ln 2 – 1/2`

`text(Let)\ \ u = x + 1 \ => \ x = u – 1`

`(du)/(dx) = 1 \ => \ du = dx`
 

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

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

`int_0^1 x/(x + 1)^2\ dx` `= int_1^2 (u – 1)/(u^2)\ du`
  `= int_1^2 1/u\ du – int_1^2 1/u^2\ du`
  `= [ln u]_1^2 + [1/u]_1^2`
  `= ln 2 – ln 1 + 1/2 – 1`
  `= ln 2 – 1/2`

 
ii.    `I_u = int_0^1 x^n/(x + 1)^2\ dx`

`u = x^n` `v prime = 1/(x + 1)^2`
`u prime = nx^(n – 1)` `v = -1/{(x + 1)}`

 

`I_n` `= [uv]_0^1 – int_0^1 u prime v\ dx`
  `= [(-x^n)/(x + 1)]_0^1 + int_0^1 (n x^(n – 1))/(x + 1)\ dx`
  `= (-1/2 – 0) + n int_0^1 (x^(n – 1) (x + 1))/(x + 1)^2\ dx`
  `= -1/2 + n int_0^1 x^n/(x + 1)^2 + (x^(n – 1))/(x + 1)^2\ dx`
  `= -1/2 + n I_n + n I_(n – 1)`
`nI_n – I_n` `= 1/2 – n I_(n – 1)`
`I_n(n – 1)` `= 1/2 – n I_(n – 1)`
`:. I_n` `= 1/(2(n – 1)) – n/(n – 1) I_(n – 1)`

 

iii.    `I_1` `= ln 2 – 1/2`
  `I_2` `= 1/(2(2 – 1)) – 2/{(2 – 1)} I_1`
    `= 1/2 – 2 I_1`
    `= 1/2 – 2 ln 2 + 1`
    `= 3/2 – 2 ln 2`

 

`:. I_3` `= 1/4 – 3/2(3/2 – 2 ln 2)`
  `= 1/4 – 9/4 + 3 ln 2`
  `= 3 ln 2 – 2`

Calculus, EXT2 C1 2016 HSC 14b

Let  `I_n = int_0^1 x^n/(x^2 + 1)^2\ dx,`  for   `n = 0, 1, 2, … .`

  1. Using a suitable substitution, show that  `I_0 = pi/8 + 1/4.`  (3 marks)

  2. Show that  `I_0 + I_2 = pi/4.`  (1 mark)

  3. Find  `I_4.`  (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `5/4 – (3pi)/8`
Show Worked Solution

i.   `I_n = int_0^1 (x^n)/((x^2 + 1)^2) dx,quadn >= 0`

`text(Let)\ \ \ x` `= tantheta`
`dx` `= sec^2theta\ d theta`

 

`text(When)\ \ \ x` `= 0,quad` `theta` `= 0`
`x` `= 1,` `theta` `= pi/4`

 

`:. I_0` `= int_0^(pi/4)(sec^2theta)/((tan^2theta + 1)^2)\ d theta`
  `= int_0^(pi/4)(sec^2theta)/(sec^4theta)\ d theta`
  `= int_0^(pi/4)cos^2theta\ d theta`
  `= int_0^(pi/4)1/2(1 + cos2theta)\ d theta`
  `= 1/2[x + 1/2sin2theta]_0^(pi/4)`
  `= 1/2[(pi/4 + 1/2 · sin\ pi/2) – 0]`
  `= pi/8 + 1/4`

 

ii.    `I_0 + I_2` `= int_0^1 1/((x^2 + 1)^2)\ dx + int_0^1 (x^2)/((x^2 + 1)^2)\ dx`
    `= int_0^1 (x^2 + 1)/((x^2 + 1)^2)\ dx`
    `= int_0^1 1/(x^2 + 1)\ dx`
    `= [tan^(−1)x]_0^1`
    `= pi/4`

 

♦ Mean mark part (iii) 49%.
iii.    `I_4` `= int_0^1 (x^4)/(x^2 + 1)\ dx`
    `= int_0^1 (x^4 – 1)/((x^2 + 1)^2)\ dx + int_0^1 1/((x^2 + 1)^2)`
    `= int_0^1 ((x^2 + 1)(x^2 – 1))/((x^2 + 1)^2)\ dx + I_0`
    `= int_0^1 (x^2 – 1)/(x^2 + 1)\ dx + I_0`
    `= int_0^1 (1 – 2/(x^2 + 1))\ dx + I_0`
    `= [x – 2tan^(−1)x]_0^1 + pi/8 + 1/4`
    `= [(1 – 2 · pi/4) – 0] + pi/8 + 1/4`
    `= 5/4 – (3pi)/8`

Calculus, EXT2 C1 2014 HSC 12d

Let  `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx`, where  `n`  is an integer and  `n ≥ 0`.

  1. Show that  `I_0 = pi/4`.  (1 mark)

  2. Show that
     
         `I_n + I_(n − 1) = 1/(2n − 1)`.  (2 marks)

  3. Hence, or otherwise, find
     
         `int_0^1 (x^4)/(x^2 + 1)\ dx`.  (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `pi/4 − 2/3`
Show Worked Solution

i.   `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx, \ \ n ≥ 0`

  `I_0` `= int_0^1 (x^0)/(x^2 + 1)\ dx`
  `= [tan^(−1) x]_0^1` 
  ` = tan^(−1) 1 − tan^(−1)0`
  `= pi/4`

 

♦ Mean mark 41%.
STRATEGY: The denominator of the proof, `(2n-1)`, should flag the potential existence of `x^(2n-2)` in the integral.
ii.  `I_(n − 1)` `= int_0^1 x^(2n-2)/(x^2 + 1)\ dx`
  `I_n` `= int_0^1 x^(2n)/(x^2 + 1)\ dx`

 

`I_n + I_(n − 1)` `= int_0^1(x^(2n))/(x^2 + 1)\ dx + int_0^1(x^(2n − 2))/(x^2 + 1)\ dx`
  `= int_0^1(x^(2n) + x^(2n − 2))/(x^2 + 1)\ dx`
  `= int_0^1(x^(2n − 2)(x^2 + 1))/(x^2 + 1)\ dx`
  `= int_0^1x^(2n − 2)\ dx`
  `= [(x^(2n − 1))/(2n − 1)]_0^1`
  `= 1/(2n − 1)`

 

iii.  `I_2` `= int_0^1(x^4)/(x^2 + 1)\ dx`
  `I_1` `= int_0^1(x^2)/(x^2 + 1)\ dx`
`I_n + I_(n − 1)` `= 1/(2n − 1)`
`I_2 + I_1` `= 1/(2(2) − 1) = 1/3\ \ …\ (1)`
`I_1 + I_0` `= 1/(2(1)-1)=1`
`I_1+pi/4` `= 1`
`:.I_1` `=1- pi/4`
`text(Substitute into (1))`
`I_2+1- pi/4` `= 1/3`
 `:.I_2` `= pi/4 − 2/3`