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Calculus, EXT2 C1 2025 HSC 13a

It is given that  \(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\, dx\).

Show that  \(\displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=k A\), where \(k\) and \(m\) are constants.   (3 marks)

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\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)
 

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\) \(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
  \(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
  \(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

 

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Show Worked Solution

\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)
 

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\) \(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
  \(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
  \(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

 

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Calculus, EXT2 C1 2024 HSC 15d

Using a suitable substitution, find  \(\displaystyle\int \dfrac{2 x^2}{\sqrt{2 x-x^2}}\, d x\).   (3 marks)

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

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  \(\displaystyle \int \dfrac{2x^2}{\sqrt{2x-x^2}}\, dx\) \(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(1-2 x+x^2)}}\, dx\)
    \(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(x-1)^2}} \, dx\)
♦ Mean mark 49%.

\(\text {Let} \ \ x-1=\sin \theta \ \Rightarrow \ x=1+\sin \theta\)

\(\dfrac{dx}{d \theta}=\cos \theta \ \Rightarrow \ dx=\cos \theta \, d \theta\)

  \(I\) \(=\displaystyle \int \frac{2(1+\sin \theta)^2}{\sqrt{1-\sin ^2 \theta}} \cdot \cos \theta \, d \theta\)
    \(=2 \displaystyle \int \frac{1+2 \sin \theta+\sin ^2 \theta}{\cos \theta} \cdot \cos \theta \, d \theta\)
    \(=2 \displaystyle \int 1+2 \sin \theta+\frac{1}{2}(1-\cos (2 \theta)) \, d \theta\)
    \(=\displaystyle \int 2+4 \sin \theta+1-\cos (2 \theta) \, d \theta\)
    \(=\displaystyle \int 3+4 \sin \theta-\cos (2 \theta) \, d \theta\)
    \(=3 \theta-4 \cos \theta-\dfrac{1}{2} \sin (2 \theta)+c\)
    \(=3 \theta-4 \cos \theta-\sin \theta \cos \theta+c\)

 

\(\Rightarrow \cos \theta=\sqrt{1^2-(x-1)^2}=\sqrt{2x-x^2}\)
 

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

Calculus, EXT2 C1 2022 SPEC1 9

Given that  `f^{\prime}(x)=\frac{\cos (2 x)}{\sin ^3(2 x)}`  and  `f((pi)/(8))=(3)/(4)`, find `f(x)`.   (4 marks)

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`f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

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`text{Let}\ \ u=sin(2x)\ \ =>\ \ {du}/{dx}=2cos(2x)`

`\int \frac{\cos (2 x)}{\sin ^3(2 x)} d x` `=\frac{1}{2} \int \frac{d u}{u^3}`  
  `= -\frac{1}{4} u^{-2}+c`  
  `=-\frac{1}{4} \ xx \frac{1}{\sin ^2(2 x)}+c`  

 
`text{When}\ \ x = pi/8:`

`-\frac{1}{4} \cdot \frac{1}{(\frac{1}{sqrt{2}})^2}+c` `=\frac{3}{4}`  
`-\frac{1}{4} \cdot 2+c` `=\frac{3}{4}`  
`\Rightarrow c` `=\frac{3}{4}+\frac{2}{4}=5/4`  

 
`:. \  f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`


♦ Mean mark 50%.

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

Show Worked Solution

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

 
\(\text{Let}\ \ u=5-4x-x^2 \)

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

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

Calculus, EXT2 C1 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 C2 2020 SPEC2 11 MC

With a suitable substitution  `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx`  can be expressed as

  1. `int_1^(1/sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  2. `int_1^(sqrt3) (1/(u - 2) - 1/(u - 1))\ du`
  3. `int_1^(sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
  4. `int_(pi/4)^(pi/3) (1/(3(u - 1)) - 1/(3(u + 2)))\ du`
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`B`

Show Worked Solution

`text(Let)\ \ u = tan(x)`

`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`

`text(When)\ `   `x = pi/3,\ u = sqrt3`
    `x = pi/4,\ u = 1`

 

`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx`

`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du`
  `= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du`
  `= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du`
  `= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du`

`=>B`

Calculus, EXT2 C1 2020 SPEC1 2

Evaluate  `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`.  (3 marks)

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`(8 sqrt 2)/3 – 10/3`

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

 

`text(When)\ \ x` `= 0,\ u = 1`
`x` `= -1,\ u = 2`

 

`int_(-1)^0 (1 + x)/sqrt(1 – x)\ dx` `= -int_2^1 (2 – u)/sqrt u\ du`
  `= int_1^2 2u^(-1/2) – u^(1/2)\ du`
  `= [4u^(1/2) – 2/3u^(3/2)]_1^2`
  `= 4 sqrt 2 – (4 sqrt 2)/3 – (4 – 2/3)`
  `= (8 sqrt 2)/3 – 10/3`

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 2019 SPEC1-N 4

Evaluate  `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`.   (3 marks)

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`log_e ((4)/(3))`

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`text(Let)\ \ u = log_e x`

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

`text(When) \ \ x = e^4 \ => \ u = 4`

`text(When) \ \ x = e^3 \ => \ u = 3`

`int_(e^3) ^(e^4) (1)/(x log_e (x))` `= int_3 ^4 (1)/(u)\ du`
  `= [ log_e u]_3 ^4`
  `= log_e 4 – log_e 3`
  `= log_e ((4)/(3))`

Calculus, EXT2 C1 2005 HSC 1a

Find  `int(cos theta)/(sin^5 theta)  d theta`   (2 marks)

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`(-1)/(4sin^4 theta ) + C`

Show Worked Solution
`text(Let) \ u` `= sin theta `
`(du)/(d theta)` `=cos theta =>  d u = cos theta \ d theta`

 

`int(cos theta)/(sin^5 theta)  d theta` `= int u^-5 d u`
  `=(-1)/(4) u ^-4 + C`
  `=(-1)/(4sin^4 theta ) + C`

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`

Show Worked Solution

`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`

Show Worked Solution

`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`

Show Worked Solution

`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`

Show Worked Solution

`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`

Show Worked Solution

`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`