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Calculus, EXT2 C1 2021 HSC 13c

  1. The integral  `I_n`  is defined by  `I_n = int_1^e (ln x)^n\ dx`  for integers  `n >= 0`.
    Show that  `I_n = e - nI_(n - 1)`  for   `n >= 1`.  (2 marks)

  2. The diagram shows two regions.

Region `A` is bounded by  `y = 1`  and  `x^2 + y^2 = 1`  between  `x = 0`  and  `x = 1`.

Region `B` is bounded by  `y = 1`  and  `y = ln x`  between  `x = 1`  and  `x = e`.
 
   
 
The volume of the solid created when the region between the curve  `y = f(x)`  and the  `x`-axis, between  `x = a`  and  `x = b`, is rotated about the `x`-axis is given by  `V = pi int_a^b [f(x)]^2\ dx`.

The volume of the solid of revolution formed when region `A` is rotated about the `x`-axis is `V_A`.

The volume of the solid of revolution formed when region `B` is rotated about the `x`-axis is `V_B`.

Using part (i), or otherwise, show that the ratio  `V_A : V_B` is `1:3`.  (4 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
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i.   `I_n = int_1^e (ln x)^n\ dx\ \ text(for)\ \ n >= 0`

`text(Show)\ \ I_n = e – nI_(n – 1)\ \ text(for)\ \ n >= 1`

`text(Integrating by parts:)`

`u` `= (ln x)^n` `u′` `= n/x (ln x)^(n – 1)`
`v` `= x` `v′` `= 1`
`I_n` `= [x(ln x)^n]_1^e – n int(ln x)^(n – 1)\ dx`
  `=[e(lne)^n-1(ln1)^n]- n I_(n – 1)`
  `= e – n I_(n – 1)`

 

ii.    `V_A` `=\ text(cylinder volume – hemisphere volume)`
    `= pi int_0^1 1\ dx – 1/2 xx 4/3 xx pir^3`
    `= pi [x]_0^1 – (2pi)/3`
    `= pi(1 – 0) – (2pi)/3`
    `= pi/3`

 

`V_B` `= pi int_1^e 1\ dx – pi int_1^e (ln x)^2\ dx`
  `= pi [x]_1^e – pi I_2`
  `= pi(e – 1) – pi(e – 2I_1)`
  `= pi(e – 1) – pi[e – 2(e – I_0)]`
  `= pi(e – 1) – pi(2I_0 – e)`
  `= pi(e – 1) – pi[2(e – 1) – e]`
  `= pi(e – 1) – pi(e – 2)`
  `= pie – pi – pie + 2pi`
  `= pi`
`:. V_A : V_B` `= pi/3 : pi`
  `= 1 : 3`