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Calculus, EXT2 C1 2023 HSC 13a

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

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\(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

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\(I\) \(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)  

 
\(\text{Let}\ \ u=5-4x-x^2 \)

\( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)

\(I\) \(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)  
  \(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  
  \(=  \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  

Calculus, EXT2 C1 2022 HSC 5 MC

If  `int_(a)^(x)f(t)dt=g(x)`, which of the following is a primitive of  `f(x)g(x)` ?

  1. `(1)/(2)[f(x)]^(2)`
  2. `(1)/(2)[f^(')(x)]^(2)`
  3. `(1)/(2)[g(x)]^(2)`
  4. `(1)/(2)[g^(')(x)]^(2)`
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`C`

Show Worked Solution
`int_a^x f(t)\ dt` `=g(x)`  
`d/dx [int_a^xf(t)\ dt]` `=g^{′}(x)`  
`f(x)` `=g^{′}(x)`  
`f(x)*g(x)` `=g^{′}(x)g(x)`  
`int f(x)*g(x)\ dx` `=int g^{′}(x)g(x)\ dx`  
  `=1/2[g(x)]^2+C`  

 
`=>C`


COMMENT: Simple examples can illustrate the logic of line 3 of the worked solution.

Calculus, EXT2 C1 2021 HSC 13b

Use an appropriate substitution to evaluate  `int_(sqrt10)^(sqrt13) x^3 sqrt{x^2 - 9}\ dx`.   (3 marks)

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`136/5`

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`int_(sqrt10)^(sqrt13) x^3 sqrt{x^2 – 9} dx`

`text{Let} \ \ u = x^2 – 9 \ => \ x^2 = u + 9`

`(du)/dx = 2x \ => \ du = 2x\ dx`
 

`text{If} \ \ x = sqrt13 \ , \ u =4`

`text{If} \ \ x = sqrt10 \ , \ u = 1`

`int_(sqrt10)^(sqrt13) x^3 sqrt{x^2 – 9}\ dx` `= int_(1)^(4) (u + 9) sqrtu * 1/2\ du`
  `= 1/2 int_(1)^(4) u^{3/2} + 9 u^{1/2}\ du`
  `= 1/2 [ 2/5 u^{5/2} + 9 * 2/3 u^{3/2}]_(1)^(4)`
  `= 1/2 [(2/5 * 4^{5/2} + 6 * 4^{3/2}) – (2/5 + 6)]`
  `= 1/2 (64/5 + 48 – 32/5)`
  `= 136/5`

Calculus, EXT2 C1 2020 HSC 10 MC

Which of the following is equal to  `int_0^(2a) f(x)\ dx`?

  1. `int_0^a f(x) - f(2a - x)\ dx`
  2. `int_0^a f(x) + f(2a -x)\ dx`
  3. `2 int_0^a f(x - a)\ dx`
  4. `int_0^a frac{1}{2} f(2x)\ dx`
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`B`

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`int_0^(2a) f(x)\ dx – int_0^a f(x)\ dx + int_a^(2a) f(x)\ dx`

♦ Mean mark 54%.

`text(Let)\ \ x = 2a – u\ \ => \ u = 2a – x`

`frac{du}{dx} = -1 \ \ => \ du = -dx`

 

`text{When}`   `x = a,` `\ u = a`
  `x= 2a,`  `\ u = 0`

 

`int_0^(2a) f(x)\ dx= int_0^a f(x)\ dx – int_a^0 f(2a – u)\ du`

`text(Use substitution for)\ \ int_a^0 f(2a – u)\ du`

`text(Let)\ \ 2a – u = 2a-x \ \ => \ x=u`

`dx/(du) = 1 \ => \ du=dx`

`text{When}`   `u = a,` `\ x = a`
  `u= 0,`  `\ x = 0`

 

`:. int_0^(2a) f(x)\ dx` `= int_0^a f(x)\ dx – int_a^0 f(2a – x)\ dx`
  `= int_0^a f(x)\ dx + int_0^a f(2a -x)\ dx`
  `= int_0^a f(x) + f(2a – x)\ dx`

 
`=> \ B`

Calculus, EXT2 C1 2017 HSC 11f

Using the substitution  `x = sin^2 theta`, or otherwise, evaluate  `int_0^(1/2) sqrt(x/(1 - x))\ dx`.  (3 marks)

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`pi/4 – 1/2`

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`x = sin^2 theta`

`dx = 2 sin theta cos theta\ d theta`

`text(If)\ \ x = 1/2, sin theta = 1/sqrt 2, \ theta = pi/4`

`text(If)\ \ x = 0, \ theta = 0`

`int_0^(1/2) sqrt(x/(1 – x))\ dx` `= int_0^(pi/4) sqrt ((sin^2 theta)/(1 – sin^2 theta)) xx 2 sin theta cos theta\ d theta`
  `= int_0^(pi/4) (sin theta)/(cos theta) xx 2 sin theta cos theta\ d theta`
  `= int_0^(pi/4) 2 sin^2 theta\ d theta`
  `= int_0^(pi/4) 1 – cos 2 theta\ d theta`
  `= [theta – 1/2 sin 2 theta]_0^(pi/4)`
  `= (pi/4 – 1/2 sin {:pi/2) – (0 – 0)`
  `= pi/4 – 1/2`

Calculus, EXT2 C1 2007 HSC 8a

  1. Using a suitable substitution, show that
     
         `int_0^a f(x)\ dx = int_0^a f(a - x)\ dx.`  (1 mark)

  2. A function  `f(x)`  has the property that  `f(x) + f(a - x) = f(a).`

     

    Using part (i), or otherwise, show that
          
         `int_0^a f(x)\ dx = a/2\ f(a).`  (2 marks)

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

i.  `text(Show)\ \ int_0^a f(x)\ dx = int_0^a f(a – x)\ dx.`

`text(Let)\ \ x = a – u,\ \ dx = -du`

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

`text(When)\ \ x = a,\ \ u = 0`
 

`:. int_0^a f(x)\ dx` `=int_a^0 f(a – u) (-du)`
  `=int_0^a f(a – u)\ du`
  `=int_0^a f(a – x)\ dx\ \ text{.. as required}`
MARKER’S COMMENT: Integrating `f(a)` was poorly done in part (ii). Pay careful attention to this in the Worked Solution.

 

ii.  `f(x) = f(a) – f(a – x)`

`int_0^a f(x)\ dx` `=int_0^a [f(a) – f(a – x)]\ dx`
  `=int_0^a f(a)\ dx – int_0^a f(a – x)\ dx`
  `=int_0^a f(a)\ dx – int_0^a f(x)\ dx`
`2 int_0^a f(x)\ dx` `=int_0^a f(a)\ dx`
  `=[f(a) xx x]_0^a`
`int_0^a f(a)\ dx` `=(f(a))/2 (a – 0)`
  `=a/2\ f(a)`

Calculus, EXT2 C1 2007 HSC 1d

Evaluate  `int_0^(3/4) x/sqrt (1 - x)\ dx.`  (4 marks)

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`5/12`

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`text(Let)\ \ u = 1 – x,\ \ du = -dx,\ \ x = 1 – u`

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

`text(When)\ \ x = 3/4,\ \ u = 1/4`

`int_0^(3/4) (x\ dx)/sqrt (1 – x)` `=int_1^(1/4) ((1 – u)(-du))/sqrt u`
  `=int_(1/4)^1 (u^(-1/2) – u^(1/2)) du`
  `=[2 u^(1/2) – 2/3 u^(3/2)]_(1/4)^1`
  `=[(2 – 2/3) – (1 – 1/12)]`
  `=5/12`

Calculus, EXT2 C1 2014 HSC 10 MC

Which integral is necessarily equal to  `int_(−a)^a f(x)\ dx`?

  1. `int_0^a f(x) − f(−x)\ dx`
  2. `int_0^a f(x) − f(a − x)\ dx`
  3. `int_0^a f(x − a) + f(−x)\ dx`
  4. `int_0^a f(x − a) + f(a − x)\ dx`
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`D`

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`int_(−a)^a f(x)\ dx= int_0^a f(x)\ dx + int_(−a)^0 f(x)\ dx`

 `text(Consider)\ \ int_0^a f(x)\ dx`

♦♦ Mean mark 29%.

`text(Let)\ u = a − x\ \  => x = a − u\ \ text(and)\ \ dx = −du`

`int_0^a f(x)\ dx` `= int_0^a f(a − u) − du`
  `=int_a^0 f(a-u)\ du`
  `= int_0^a f(a − x)\ dx`

 

`text(Consider)\ \ int_(−a)^0 f(x)\ dx`

`text(Let)\ u = x + a\ \  => x = u − a\ \ text(and)\ \ dx = du`

 `int_(−a)^0 f(x)\ dx` `= int_0^a f(u − a)\ du`
  `= int_0^a f(x − a)\ dx`
`:.int_(−a)^a f(x)\ dx` `= int_0^a f(x − a)+  f(a − x)\ dx`

 

`=> D`

Calculus, EXT2 C1 2009 HSC 1e

Evaluate  `int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx.`   (4 marks)

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`sqrt 2 – (2 sqrt 3)/3`

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

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

`text(When)\ \ x = sqrt 3,\ \ theta = pi/3`

`int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx` `= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sqrt (1 + tan^2 theta))\ d theta`
  `= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sec theta)\ d theta`
  `= int_(pi/4)^(pi/3) (sec theta)/(tan^2 theta)\ d theta`
  `= int_(pi/4)^(pi/3) (cos^2 theta)/(cos theta sin^2 theta)\ d theta`
  `= int_(pi/4)^(pi/3) (cos theta\ d theta)/(sin^2 theta)`
  `= [-1/(sin theta)]_(pi/4)^(pi/3)`
  `= (-2/sqrt 3 + sqrt 2)`
  `= sqrt 2 – (2 sqrt 3)/3`

Calculus, EXT2 C1 2010 HSC 1e

Find  `int (dx)/(1 + sqrtx)`.   (3 marks)

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`2sqrtx − 2log_e\ (1 + sqrtx) + c_1\ \ text(or)`

`2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

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`text(Solution 1)`

`text(Let)\ \ u = sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2u\ du`
`int (dx)/(1 + sqrtx)` `=int (2u\ du)/(1 + u)`
  `=int((2+2u-2)/(1+u))\ du`
  `=int (2 − 2/(1 + u))\ du`
  `=2u − 2log_e(1 + u) + c_1`
  `=2sqrtx − 2log_e\ (1 + sqrtx) + c_1`

 

`text(Alternative Solution)`

`text(Let)\ \ u = 1+sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2(u-1)\ du`
`int (dx)/(1 + sqrtx)` `=2int (u-1)/u\ du`
  `=2 int(1 − 1/u)du`
  `=2u − 2log_e\ |\ u\ | + c_2`
  `=2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

 

`text(NB. These solutions are equivalent by making)\ \ c_1=c_2+2`

Calculus, EXT2 C1 2011 HSC 1b

Evaluate  `int_0^3 x sqrt (x + 1)\ dx.`  (3 marks)

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`116/15`

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`text(Let)\ \ u = 1 + x,\ \ \ du = dx`

`text(When)\ \ x = 0,` `\ \ \ u = 1`
`x = 3,` `\ \ \ u = 4`

 

`int_0^3 x sqrt (1 + x)\ dx` `= int_1^4 (u – 1) sqrt u\ du`
  `= int_1^4 (u^(3/2) – u^(1/2))\ du`
  `= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^4`
  `= (2/5 xx 32 – 2/3 xx 8) – (2/5 – 2/3)`
  `= 116/15`

Calculus, EXT2 C1 2013 HSC 14a

The diagram shows the graph  `y = ln x.`
 


 

By comparing relevant areas in the diagram, or otherwise, show that

`ln t > 2 ((t - 1)/(t + 1))`, for `t > 1.`  (3 marks)

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

Show Worked Solution
`text(Area under curve)` `>\ text(Area of triangle)`
`int_1^t ln x\ dx` `> 1/2 xx (t – 1)ln t,\ \ \ t > 1`
`underbrace{int_1^t 1\ln x\ dx}_text(integration by parts)` `> 1/2 xx (t – 1)ln t`
`[x ln x]_1^t – int_1^t x * 1/x\ dx` `> ((t – 1)ln t)/2`
`(tlnt – ln 1) – [x]_1^t` `> ((t – 1)ln t)/2`
`t ln t – (t – 1)` `> ((t – 1) ln t)/2`
`2t ln t – 2t + 2` `> t ln t – ln t`
`t ln t + ln t` `> 2(t – 1)`
`(t + 1) ln t` `> 2 (t – 1)`
`ln t` `> 2 ((t – 1)/(t + 1)),\ \ \ t > 1`

Calculus, EXT2 C1 2013 HSC 11d

Evaluate  `int_0^1 x^3 sqrt(1 - x^2)\ dx.`  (3 marks)

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`2/15`

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STRATEGY: The choice of `u=1-x^2` or `u^2=1-x^2` provided much less calculation than `u=x^2`. Take note!

`text(Let)\ \ u = 1 – x^2\ ,\ \ du = -2x\ dx`

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

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

`int_0^1 x^3 sqrt(1 – x^2)\ dx` `= int_0^1 x^2 sqrt (1 – x^2) \ x\ dx`
  `= int_1^0 (1 – u) sqrt u xx ((-du)/2)`
  `= 1/2 int_0^1 (u^(1/2) – u^(3/2))\du`
  `= 1/2 [2/3 u^(3/2) – 2/5 u^(5/2)]_1^0`
  `= 1/2 [(2/3 – 2/5)-0]`
  `= 2/15`