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Calculus, EXT2 C1 2024 HSC 15b

Let  \(I_n=\displaystyle\int_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, d x\), where  \(n \geq 0\).

Show that \((2 n+4) I_n=a(2 n+1) I_{n-1}\), for  \(n>0\).   (3 marks)

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\(I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,d x \ \ \text{where}\ \ n \geqslant 0\)

\(\text{Show}\ \ (2 n+4) I_n=a(2 n+1) I_{n-1}\ \ \text{for}\ \ n>0\)

\(I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\)

\(\text{Using integration by parts: }\)

\(\begin{array}{ll}u=x^{n+\frac{1}{2}} & v^{\prime}=(a-x)^{\frac{1}{2}} \\ u^{\prime}=\left(n+\frac{1}{2}\right) x^{n-\frac{1}{2}} & v=-\dfrac{2}{3}(a-x)^{\frac{3}{2}}\end{array}\)

  \(I_n\) \(=\left[u v\right]_0^a-\displaystyle{\int}_0^a u^{\prime} v\, dx\)
    \(=\underbrace{\left[-\dfrac{2}{3} x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\right]_0^a}_{=0}+\dfrac{2}{3}\left(n+\frac{1}{2}\right) \displaystyle{\int}_0^a x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\, dx\)
    \(=\dfrac{2 n+1}{3} \displaystyle{\int}_0^a a x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}-x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, dx\)
  \(3I_n\) \(=(2 n+1)\left[a I_{n-1}-I_n\right]\)

 

  \(3 I_n+(2 n+1) I_n\) \(=a(2 n+1) I_{n-1}\)
  \((2 n+4) I_n\) \(=a(2 n+1) I_{n-1}\)

Show Worked Solution

\(I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,d x \ \ \text{where}\ \ n \geqslant 0\)

\(\text{Show}\ \ (2 n+4) I_n=a(2 n+1) I_{n-1}\ \ \text{for}\ \ n>0\)

\(I_n=\displaystyle{\int}_0^a x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\)

\(\text{Using integration by parts: }\)

\(\begin{array}{ll}u=x^{n+\frac{1}{2}} & v^{\prime}=(a-x)^{\frac{1}{2}} \\ u^{\prime}=\left(n+\frac{1}{2}\right) x^{n-\frac{1}{2}} & v=-\dfrac{2}{3}(a-x)^{\frac{3}{2}}\end{array}\)

  \(I_n\) \(=\left[u v\right]_0^a-\displaystyle{\int}_0^a u^{\prime} v\, dx\)
    \(=\underbrace{\left[-\dfrac{2}{3} x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\right]_0^a}_{=0}+\dfrac{2}{3}\left(n+\frac{1}{2}\right) \displaystyle{\int}_0^a x^{n-\frac{1}{2}}(a-x)^{\frac{3}{2}}\, dx\)
    \(=\dfrac{2 n+1}{3} \displaystyle{\int}_0^a a x^{n-\frac{1}{2}}(a-x)^{\frac{1}{2}}-x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\, dx\)
  \(3I_n\) \(=(2 n+1)\left[a I_{n-1}-I_n\right]\)

  \(3 I_n+(2 n+1) I_n\) \(=a(2 n+1) I_{n-1}\)
  \((2 n+4) I_n\) \(=a(2 n+1) I_{n-1}\)

Calculus, EXT2 C1 2018 HSC 14c

Let  `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx`  for  `n = 0, 1, 2…`

  1.  Show that, for  `n >= 1`,
     
         `I_n = (−6n)/(3 + 2n) I_(n - 1)`.   (3 marks)

  2.  Find the value of  `I_2`.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `(144sqrt3)/35`
Show Worked Solution

i.   `I_n = int_(−3)^0 x^n sqrt(x + 3)\ dx\ \ text(for)\ \ n = 0, 1, 2…`

`u` `= x^n` `vprime` `= (x + 3)^(1/2)`
`uprime` `= nx^(n – 1)` `v` `= 2/3 (x + 3)^(3/2)`

 

`I_n` `= [2/3 x^n(x + 3)^(3/2)]_(−3)^0 – 2/3int_(−3)^0  nx^(n – 1)(x + 3)sqrt(x + 3)\ dx`
`I_n` `= 0 – (2n)/3 int_(−3)^0 x^nsqrt(x + 3) + 3x^(n – 1)sqrt(x + 3)\ dx`
`I_n` `= −(2n)/3 (I_n + 3I_(n-1))`
`I_n + (2n)/3 I_n` `= −2nI_(n – 1)`
`I_n(1 + (2n)/3)` `= −2nI_(n – 1)`
`I_n((3 + 2n)/3)` `= −2nI_(n – 1)`
`:. I_n` `= (−6n)/(3 + 2n) I_(n – 1)\ \ \ text(… as required)`

 

ii.    `I_2` `= −12/7 I_1`
    `= −12/7 xx −6/5 I_0`
    `= 72/35 int_(−3)^0 (x + 3)^(1/2)dx`
    `= 72/35 xx 2/3 [(x + 3)^(3/2)]_(−3)^0`
    `= 48/35 (3^(3/2))`
    `= (144sqrt3)/35`

Calculus, EXT2 C1 2017 HSC 15a

Let  `I_n = int_0^1 x^n sqrt (1 - x^2)\ dx`, for  `n = 0, 1, 2, …`

  1. Find the value of `I_1`.  (1 mark)

  2. Using integration by parts, or otherwise, show that for  `n >= 2`
     
         `I_n = ((n - 1)/(n + 2)) I_(n-2)`.  (3 marks)

  3. Find the value of  `I_5`.  (1 mark)

Show Answers Only
  1. `1/3`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `8/105`
Show Worked Solution
i.   `I_1` `= int_0^1 x^1 sqrt (1-x^2)\ dx`
    `= -1/2 int_0^1-2x sqrt (1-x^2)\ dx`
    `= -1/2 [2/3 (1-x^2)^(3/2)]_0^1`
    `= -1/2 [0-2/3 (1)]`
    `= 1/3`

 

ii.  `I_n = int_0^1 x^n sqrt (1-x^2)\ dx`

♦ Mean mark 48%.

`text(Integrating by parts:)`

`u` `= x^(n-1)`    `dv` `= x sqrt (1-x^2)\ dx`
`du` `= (n-1) x^(n – 2)\ dx` `v` `= -1/3 (1-x^2)^(3/2)`
`I_n` `= [x^(n-1) xx -1/3 (1-x^2)^(3/2)]_0^1 + 1/3 int_0^1 (1-x^2)^(3/2)(n-1) x^(n-2)\ dx`
  `= 0 + (n-1)/3 int_0^1 x^(n-2) (1-x^2) · sqrt(1-x^2)\ dx`
  `= (n-1)/3 int_0^1 x^(n-2) sqrt(1-x^2)\ dx-(n-1)/3 int_0^1 x^n sqrt (1-x^2)\ dx`
  `= (n-1)/3 · I_(n-2)-(n-1)/3 · I_n`

 

`:. I_n + (n-1)/3 · I_n` `= (n-1)/3 · I_(n-2)`
`I_n (1 + (n-1)/3)` `= (n-1)/3 · I_(n-2)`
`I_n ((n + 2)/3)` `= (n-1)/3 · I_(n-2)`
`:. I_n` `= ((n-1)/(n + 2)) · I_(n-2)\ text(… as required.)`

 

iii.   `I_5` `= 4/7 xx I_3`
    `= 4/7 xx 2/5 xx I_1`
    `= 4/7 xx 2/5 xx 1/3\ text{(part (i))}`
    `= 8/105`

Calculus, EXT2 C1 2011 HSC 8a

For every integer `m >= 0`  let

`I_m = int_0^1 x^m (x^2 - 1)^5\ dx.`
 

Prove that for  `m >= 2`

`I_m = (m - 1)/(m + 11) \ I_(m - 2).`   (3 marks)

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

Show Worked Solution

`I_m = int_0^1 x^m(x^2 – 1)^5 dx`

`u` `=x^(m – 1)\ \ \ \ \ \ ` `\ \ \ \ u′` `=(m – 1)x^(m – 2)`
`v′` `=x(x^2 – 1)^5` `v` `=1/12 (x^2 – 1)^6`
♦♦♦ Mean mark 24%.

 

`I_m` `=[x^(m-1) xx 1/12(x^2 – 1)^6]_0^1 – int_0^1 1/12(x^2 – 1)^6 xx (m – 1)x^(m – 2) dx`
  `=[0 – 0] – (m – 1)/12 int_0^1 x^(m – 2) (x^2 – 1)^6 dx`
  `=-(m – 1)/12 int_0^1 x^(m – 2) (x^2 – 1)^5 (x^2 – 1) dx`
  `=-(m – 1)/12 int_0^1 x^m (x^2 – 1)^5 dx + (m – 1)/12 int _0^1 x^(m – 2) (x^2 – 1)^5 dx`
  `=-(m – 1)/12 I_m + (m – 1)/12 I_(m – 2)`

 

`12I_m + mI_m – I_m` `= (m – 1)I_(m – 2)`
`(m + 11)I_m` ` = (m – 1)I_(m-2)`
`:.I_m` `= (m – 1)/(m + 11) I_(m – 2)`

Calculus, EXT2 C1 2013 HSC 13a

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

  1. Show that
     
            `I_n = n/(n + 1) I_(n-2)`  for every integer  `n>= 2.`  (3 marks)

  2. Evaluate  `I_5.`  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `(5 pi)/32`
Show Worked Solution

i.  `I_n = int_0^1 (1 – x^2)^(n/2)\ dx`

♦ Mean mark 46%.
STRATEGY TIP: Attempting to prove this relationship using induction proved unsuccessful and time consuming for some students.
`text(Let)\ \ u` `= (1 – x^2)^(n/2)`
`u′` `= n/2 (1 – x^2)^(n/2 – 1) (-2x)`
  `= -nx (1 – x^2)^(n/2 -1)`
`v` `=x`
`v′` `=1`

 

`I_n` `= [x (1 – x^2)^(n/2)]_0^1 – int_0^1 – nx (1 – x^2)^(n/2 – 1)x\ dx`
  `= 0 + n int_0^1 x^2 (1 – x^2)^(n/2 – 1)\ dx`
  `= n int_0^1 (x^2 – 1 + 1) (1 – x^2)^(n/2 – 1)\ dx`
  `= n int_0^1 1 (1 – x^2)^(n/2 -1)\ dx – n int_0^1 (1 – x^2) (1 – x^2)^(n/2 – 1)\ dx`
  `= n int_0^1 (1 – x^2)^((n-2)/2)\ dx – n int_0^1 (1 – x^2)^(n/2)\ dx`
  `= nI_(n – 2) – nI_n`

 

`:.(n + 1) I_n` `= nI_(n-2)`
`I_n` `= n/(n + 1) I_(n-2)\ \ \ \ text(where)\ \ n >= 2`
STRATEGY TIP: Realising that the integral `I_1` is equal to the area of a quarter unit circle saved valuable time.

 

ii.  `I_5 = 5/6 I_3 = 5/6 xx 3/4\ I_1`

`I_1` `= int_0^1 (1 – x^2)^(1/2)\ dx`
  `= 1/4 xx pi xx 1^2\ \ \ \ text{(1st quadrant of a circle, radius 1)}`
  `= pi/4`
`I_5` `= 5/6 xx 3/4 xx pi/4`
  `= (5 pi)/32`