SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, EXT1 C2 EQ-Bank 30

Solve \( \displaystyle\int_0^1 \dfrac{1}{(1+x^2)^\tfrac{3}{2}}\, dx \)  using the substitution  \(x=\tan \theta\).   (3 marks)

Show Answers Only

\(\dfrac{1}{\sqrt{2}}\)

Show Worked Solution

\(\text {Let} \ \ x=\tan \theta\)

\(\dfrac{d x}{d \theta}=\sec ^2 \theta \ \Rightarrow \ d x=\sec ^2 \theta \, d \theta\)

\(\text{Also,} \ \ 1+x^2=1+\tan ^2 \theta=\sec ^2 \theta\)

\(\left(1+x^2\right)^{\tfrac{3}{2}}=(\sec ^2 \theta)^{\tfrac{3}{2}}=\sec ^3 \theta\)
 

\(\text{Adjust limits:}\)

\(x=0 \Rightarrow \theta=0, \ \ x=1 \Rightarrow \theta=\dfrac{\pi}{4}\)

\(\displaystyle\int_0^1 \dfrac{1}{\left(1+x^2\right)^{\tfrac{3}{2}}} \, d x\) \(=\displaystyle\int_0^{\tfrac{\pi}{4}} \frac{\sec ^2 \theta}{\sec ^3 \theta} \, d \theta\)
  \(=\displaystyle \int_0^{\tfrac{\pi}{4}} \cos \theta \, d \theta\)
  \(=\Big[\sin \theta\Big]_0^{\tfrac{\pi}{4}}\)
  \(=\sin \dfrac{\pi}{4}-\sin 0\)
  \(=\dfrac{1}{\sqrt{2}}\)

Calculus, EXT1 C2 EQ-Bank 18

Using the substitution  \(u=\sqrt{x}+1\), evaluate  \(\displaystyle \int_4^9 \dfrac{e^{\sqrt{x}+1}}{\sqrt{x}}\, d x\).   (3 marks)

Show Answers Only

\(2 e^3(e-1)\)

Show Worked Solution

\(\text{Let} \ \ u=\sqrt{x}+1\)

\(\dfrac{d u}{d x}=\dfrac{1}{2 \sqrt{x}}\ \ \Rightarrow \ \ 2\, d u=\dfrac{1}{\sqrt{x}}\, d x\)

\(\text{When} \ \ x=9, u=4\)

\(\text{When} \ \  x=4, x=3\)

\(\displaystyle\int_4^9 \dfrac{e^{\sqrt{x}+1}}{\sqrt{x}}\, d x\) \(=2 \displaystyle \int_3^4 e^u\, d u\)
  \(=2\big[e^u \big]_3^4\)
  \(=2 e^4-2 e^3\)
  \(=2 e^3(e-1)\)

Calculus, EXT1 C2 2025 SPEC2 7 MC

Using the substitution  \(u=\cos (\theta), \ \dfrac{1}{2} \displaystyle \int_0^{\tfrac{\pi}{2}} \dfrac{\sin (2 \theta)}{1+\cos (\theta)} \, d \theta\)  can be expressed as

  1. \(\displaystyle\int_0^{\tfrac{\pi}{2}} u \sqrt{\frac{1-u}{1+u}}\ d u\)
  2. \(\displaystyle\int_0^1\left(1+\frac{1}{1+u}\right) d u\)
  3. \(\displaystyle\int_0^{\tfrac{\pi}{2}}\left(1-\frac{1}{1+u}\right) d u\)
  4. \(\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) d u\)
Show Answers Only

\(D\)

Show Worked Solution

\(u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta\)

\(\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0\)

\(\text{When}\ \  \theta=0, u=1\)

\(I\) \(=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta\)
  \(=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta\)
  \(=\displaystyle \int_1^0-\frac{u}{1+u} \, d u\)
  \(=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u\)
  \(=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u\)

 

\(\Rightarrow D\)

Calculus, EXT1 C2 2025 HSC 3 MC

Consider the integral  \(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\).

The substitution  \(x=5 \sin \theta\)  is applied.

Which of the following is obtained?

  1. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec} \theta \, d \theta\)
  2. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \,  d \theta\)
  3. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec}^2 \theta \, d \theta\)
  4. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec ^2 \theta \, d \theta\)
Show Answers Only

\(\Rightarrow B\)

Show Worked Solution

\(x=5\, \sin \theta\)

\(\dfrac{dx}{d \theta}=5\, \cos \theta \ \Rightarrow \ dx=5\, \cos \theta \, d \theta\)

\(\text{When} \ \ x=\dfrac{5}{2} \ \Rightarrow \ \sin\, \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text{When} \ \ x=-\dfrac{5}{2} \ \Rightarrow \ \sin \theta=-\dfrac{1}{2} \ \Rightarrow \ \theta=-\dfrac{\pi}{6}\)

\(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\) \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\left(\dfrac{1}{25-25\, \sin ^2 \theta}\right) \cdot 5\  \cos \theta \, d \theta\)
  \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{5\, \cos \theta}{25\, \cos ^2 \theta} \,  d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{1}{\cos \theta} \, d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \, d \theta\)

\(\Rightarrow B\)

Calculus, EXT1 C2 EQ-Bank 28

Using the substitution  \(x=\cos 2 \theta\),  show

\(\displaystyle \int \sqrt{\frac{1-x}{1+x}}\,dx=\sqrt{1-x^2}-\cos ^{-1} x+c\)    (4 marks)

Show Answers Only

\(\text{See worked solutions}\)

Show Worked Solution

\(x=\cos 2 \theta\)

\(\dfrac{dx}{d \theta}=-2 \sin 2 \theta \ \Rightarrow\ \ dx=-2 \sin 2 \theta\, d \theta\)

\(\displaystyle \int \sqrt{\frac{1-x}{1+x}} \, dx\) \(=\displaystyle\int \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \times -2 \sin 2 \theta\, d \theta\)
  \(=\displaystyle\int \sqrt{\frac{1-\left(2 \cos ^2 \theta-1\right)}{1+\left(2 \cos ^2 \theta-1\right)}} \times -4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int \sqrt{\frac{2\left(1-\cos ^2 \theta\right)}{2 \cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int\sqrt{\dfrac{\sin ^2 \theta}{\cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int-4 \sin ^2 \theta \, d \theta\)
  \(=-4 \displaystyle \int \frac{1-\cos 2 \theta}{2} \,d \theta\)
  \(=-2 \displaystyle \int 1-\cos 2 \theta \, d \theta\)
  \(=-2 \displaystyle \int 1\, d \theta+2 \int \cos 2 \theta \, d \theta\)
  \(=-2 \theta+\sin 2 \theta+c\)
  \(=-\cos ^{-1} x+\sqrt{1-\cos ^2 2 \theta}+c \quad \text {(note:}\ \  x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x \text{)}\)
  \(=-\cos ^{-1} x+\sqrt{1-x^2}+c\)

Calculus, EXT1 C2 EQ-Bank 20

Use the substitution  \(u=\cos\,x\)  to evaluate

\(\displaystyle {\int}_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin x\, dx\).   (3 marks)

Show Answers Only

\(1-\dfrac{1}{e}\)

Show Worked Solution

\(u=\cos x\)

\(\dfrac{d u}{d x}=-\sin\,x\  \Rightarrow\  du=-\sin\,x\,dx\)

\(\text {When} \ \ x=\pi, u=-1\)

\(\text{When} \ \ x=\dfrac{\pi}{2}, u=0\)

\(\displaystyle \int_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin\,x\,d x\) \(=\displaystyle -\int_0^{-1} e^u\,d u\)
  \(=\left[-e^u\right]_0^{-1}\)
  \(=-e^{-1}+e^0\)
  \(=1-\dfrac{1}{e}\)

Calculus, EXT1 C2 2024 HSC 13d

Using the substitution  \(u=e^x+2 e^{-x}\),  and considering \(u^2\), find  \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\).   (3 marks)

Show Answers Only

\(\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

Show Worked Solution

\(u=e^x+2 e^{-x} \ \Rightarrow \ u^2=\left(e^x+2 e^{-x}\right)^2=e^{2 x}+4+4 e^{-2 x}\)

\(\dfrac{du}{dx}=e^x-2 e^{-x} \ \Rightarrow \ du=\left(e^x-2 e^{-x}\right)\, d x\)

Mean mark 55%.
  \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\)
    \(=\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}} \times \frac{e^{-2 x}}{e^{-2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4 e^{-2 x}+8+e^{2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4+\left(e^{2 x}+4+4 e^{-2 x}\right)}\, d x\)
    \(=\displaystyle \int \frac{1}{4+u^2}\, d u\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{u}{2}\right)+c\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

Calculus, EXT1 C2 2024 HSC 11c

Using the substitution  \(u=x-1\), find  \(\displaystyle \int x \sqrt{x-1}\, d x\).   (3 marks)

Show Answers Only

\(\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\)

Show Worked Solution

\(\displaystyle \int x \sqrt{x-1}\, d x \)

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

 

  \(\displaystyle\int(u+1) \sqrt{u+1-1}\, d u\) \(=\displaystyle{\int}(u+1) \sqrt{u} \, d u\)
    \(=\displaystyle{\int} u^{\frac{3}{2}}+u^{\frac{1}{2}}\, d u\)
    \(=\dfrac{2}{5} u^{\frac{5}{2}}+\dfrac{2}{3} u^{\frac{3}{2}}+c\)
    \(=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+\dfrac{2}{3}(x-1)^{\frac{3}{2}}+c\)

Calculus, EXT1 C2 2023 HSC 12a

Evaluate \(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) using the substitution  \(u=x-3\).   (3 marks) 

Show Answers Only

\(\dfrac{56}{15}\)

Show Worked Solution

\(u=x-3\ \ \Rightarrow \ x=u+3 \)

\(\dfrac{du}{dx}=1\ \ \Rightarrow \ du=dx \)

\(\text{When}\ \ x=4, u=1 \)

\(\text{When}\ \ x=3, u=0 \)

\(\displaystyle \int_3^4(x+2) \sqrt{x-3}\ dx\) \(=\displaystyle \int_0^1(u+5) \sqrt{u}\ du\)  
  \(=\displaystyle \int_0^1 u^\frac{3}{2} +5u^\frac{1}{2}\ du\)  
  \(=\Big{[}\dfrac{2}{5} \times u^\frac{5}{2} + \dfrac{2}{3} \times 5u^\frac{3}{2}\Big{]}_0^1 \)  
  \(=\Big{[} \Big{(}\dfrac{2}{5} + \dfrac{10}{3}\Big{)}-0\Big{]}\)  
  \(=\dfrac{56}{15}\)  

Calculus, EXT1 C2 2022 HSC 11b

Find the exact value of  `int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx`  using the substitution `u=x^(2)+4`.  (3 marks)

Show Answers Only

`sqrt5-2`

Show Worked Solution

`u=x^(2)+4`

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

`text{At}\ \ x=1,\ \ u=5`

`text{At}\ \ x=0,\ \ u=4`

`int_(0)^(1)(x)/(sqrt(x^(2)+4))\ dx` `=1/2 int_(4)^(5)(1)/(sqrt(u))\ du`  
  `=1/2[2sqrtu]_4^5`  
  `=[sqrtu]_4^5`  
  `=sqrt5-2`  

Calculus, EXT1 C2 2021 HSC 11c

Use the substitution  `u = x + 1`  to find  `int xsqrt(x + 1)\ dx`.  (3 marks)

Show Answers Only

`2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

Show Worked Solution

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

`int x sqrt(x – 1)\ dx` `= int (u – 1) sqrtu\ du`
  `= int u^(3/2) – u^(1/2)\ du`
  `= 2/5 u^(5/2) – 2/3 u^(3/2) + c`
  `= 2/5 (x + 1)^(5/2) – 2/3(x + 1)^(3/2) + c`

Calculus, EXT1 C2 2020 SPEC1 2

Evaluate  `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`, using the substitution  `u=1-x`.  (3 marks)

Show Answers Only

`(8 sqrt 2)/3 – 10/3`

Show Worked Solution
`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, EXT1 C2 2020 HSC 13a

  1. Find  `d/(d theta) (sin^3 theta)`.  (1 mark)

  2. Use the substitution  `x = tan theta`  to evaluate  `int_0^1 (x^2)/(1 + x^2)^(5/2)\ dx`.  (4 marks)

Show Answers Only
  1. `3 cos theta sin^2 theta`
  2. `sqrt2/12`
Show Worked Solution

i.   `d/(d theta) (sin^3 theta) = 3 cos theta sin^2 theta`

 

ii.   `text(Let)\ x = tan theta`

`(dx)/(d theta) = sec^2 theta \ => \ dx  = sec^2 theta\ d theta`

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

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

`int_0^1 (x^2)/(1 + x^2)^(5/2) dx` `= int_0^(pi/4) (tan^2 theta)/((1 + tan^2 theta)^(5/2)) xx sec^2 theta\ d theta`
  `= int_0^(pi/4) (tan^2 theta)/((sec^2 theta)^(5/2)) xx sec^2 theta\ d theta`
  `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/((sec^2 theta)^(3/2))\ d theta`
  `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/(sec^3 theta)\ d theta`
  `= int_0^(pi/4) sin^2 theta cos theta\ d theta`
  `= 1/3[sin^3 theta]_0^(pi/4)`
  `= 1/3(sin^3\ pi/4 – 0)`
  `= 1/3 (1/sqrt2)^3`
  `= 1/(6sqrt2)`
  `= sqrt2/12`

Calculus, EXT1 C2 EQ-Bank 5 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

  1. `1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
  2. `2 int_3^11 (u^(3/2) + u^(1/2))\ du`
  3. `2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
  4. `1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

`(du)/(dx) = 2 \ => \ dx = 1/2 du`

`text(When)\ \ x = 5, \ u = 11`

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

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

 
`=>D`

Calculus, EXT1 C2 2019 HSC 13a

Use the substitution  `u = cos^2 x`  to evaluate  `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`.  (3 marks)

Show Answers Only

`ln {:10/9`

Show Worked Solution
`u` `= cos^2 x`
`(du)/(dx)` `= -2 sin x cos x`
  `= -sin 2x`
`du` `= -sin 2x\ dx`

 
`text(When)\ \ x = pi/4,\ \ u = 1/2`

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

`:. int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx` `= -int_1^(1/2) (du)/(4 + u)`
  `= -[ln (4 + u)]_1^(1/2)`
  `= -(ln 4.5 – ln 5)`
  `= -ln {:9/10`
  `= ln {:10/9`

Calculus, EXT1 C2 2018 HSC 11f

Evaluate  `int_-3^0 x/sqrt(1 - x) dx`, using the substitution  `u = 1 - x`.  (3 marks)

Show Answers Only

`- 8/3`

Show Worked Solution
`u` `= 1 – x\ \ => x=1-u`
`(du)/dx` `= -1`
`dx` `= -du`

 
`text(When)\ \ x=0, \ u=1`

`text(When)\ \ x=-3, \ u=4`
 
`int_-3^0 x/sqrt (1- x)\ dx`

  `= -int_4^1 (1 – u)/sqrt(u)\ du`
  `= – int_4^1 u^(- 1/2) – u^(1/2)\ du`
  `= – [2 u^(1/2) – 2/3 u^(3/2)]_4^1`
  `= -[(2-2/3) – (2 sqrt 4 – 2/3 (sqrt 4)^3)`
  `= -[4/3 – (4-16/3)]`
  `= -8/3`

Calculus, EXT1 C2 2017 HSC 11e

Evaluate  `int_0^3 x/sqrt(x + 1)\ dx`, using the substitution  `x = u^2 - 1`.  (3 marks)

Show Answers Only

`8/3`

Show Worked Solution
`x` `=u^2 – 1`
`u^2` `= x + 1`
`u` `= sqrt(x +1)`
`du` `= 1/(2sqrt(x + 1))\ dx`

 

`text(If)qquadx` `= 3,` `u` `= 2`
`x` `= 0,` `u` `= 1`
`:. int_0^3 x/sqrt(x + 1)\ dx` `= 2 int_1^2 u^2 – 1\ du`
  `= 2[(u^3)/3 – u]_1^2`
  `= 2[(8/3 – 2) – (1/3 – 1)]`
  `= 2(2/3 + 2/3)`
  `= 8/3`

Calculus, EXT1 C2 2016 HSC 11b

Use the substitution  `u = x - 4`  to find  `int xsqrt(x - 4)\ dx`.  (3 marks)

Show Answers Only

`2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

Show Worked Solution

`u = x – 4\ \ => \ x = u + 4`

`(du)/(dx) = 1\ \ => \ dx = du`
 

`:. int x sqrt (x – 4)\ dx`

`= int (u + 4) · u^(1/2)\ du`

`= int u^(3/2) + 4u^(1/2)\ du`

`= 2/5 u^(5/2) + 4 · 2/3 u^(3/2) + c`

`= 2/5 (x – 4)^(5/2) + 8/3 (x – 4)^(3/2) + c`

Calculus, EXT1 C2 2007 HSC 1e

Use the substitution  `u = 25 - x^2`  to evaluate  `int_3^4 (2x)/(sqrt(25 - x^2))\ dx`.  (3 marks)

Show Answers Only

`2`

Show Worked Solution
`u` `= 25 − x^2`
`(du)/(dx)` `= -2 x`
`du`  `= -2x\ dx`
`text(If)`   `x = 4,`   `u = 9`
    `x = 3,`   `u = 16`

 

`:. int_3^4 (2x)/(sqrt(25 − x^2))\ dx`

MARKER’S COMMENT: A “significant number” of students put the integral limits in the wrong order.

`= − int_16^9 u^(−1/2) du`

`= − [1/(1/2) u^(1/2)]_16^9`

`= − [2sqrtu]_16^9`

`= − [2sqrt9 − 2sqrt16]`

`= − [6 − 8]`

`= 2`

Calculus, EXT1 C2 2004 HSC 1e

Use the substitution  `u = x − 3` to evaluate

`int_3^4 xsqrt(x − 3)\ dx.` (3 marks)

Show Answers Only

`2 2/5`

Show Worked Solution

`text(Let)\ \ u = x − 3`

`=> x = u + 3`

`(du)/dx = 1`

`=> dx = du`

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

 
`int_3^4 xsqrt(x − 3\ dx)`

`= int_0^1(u + 3)\ u^(1/2)\ du`

`= int_0^1u^(3/2) + 3u^(1/2)\ du`

`=[2/5u^(5/2) + 3 xx 2/3u^(3/2)]_0^1`

`= [2/5u^(5/2) + 2u^(3/2)]_0^1`

`= [(2/5 + 2) − 0]`

`= 2 2/5`

Calculus, EXT1 C2 2006 HSC 1b

Using the substitution  `u =x^4 + 8`, or otherwise, find

`int x^3 sqrt (x^4 + 8)\ dx.`  (3 marks)

Show Answers Only

`1/6 (x^4 + 8)^(3/2) + c`

Show Worked Solution

`u = x^4 + 8`

`(du)/(dx)` `= 4x^3`
`1/4 du` `= x^3 dx` 

 
`:. int x^3 sqrt (x^4 + 8)\ dx`

`= int u^(1/2) *1/4 * du`

`= 1/4 int u^(1/2) du`

`= 1/4 * 2/3 * u^(3/2) + c`

`= 1/6 u^(3/2) + c`

`= 1/6 (x^4 + 8)^(3/2) + c`

Calculus, EXT1 C2 2005 HSC 1d

Using the substitution  `u = 2x^2 + 1`, or otherwise, find  `int x (2x^2 + 1)^(5/4)\ dx.`  (3 marks)

Show Answers Only

`1/9 (2x^2 + 1)^(9/4) + c`

Show Worked Solution

 `u = 2x^2 + 1`

`(du)/(dx)` `= 4x`
`du` `= 4x\ dx`
`dx` `=(du)/(4x)` 

 

`:.\ int x (2x^2 + 1)^(5/4)\ dx`

`= int x * u^(5/4) * (du)/(4x)`

`= 1/4 int u^(5/4)\ du`

`= 1/4 * 4/9\ u^(9/4) + c`

`= 1/9\ u^(9/4) + c`

`= 1/9 (2x^2 + 1)^(9/4) + c`

Calculus, EXT1 C2 2015 HSC 11e

Use the substitution  `u = 2x - 1`  to evaluate  `int_1^2 x/((2x - 1)^2)\ dx`.  (3 marks)

Show Answers Only

`1/4(ln 3 + 2/3)`

Show Worked Solution

`u = 2x − 1`

`⇒ 2x` `= u + 1`
 `x` `= 1/2(u + 1)`
`(du)/(dx)` `= 2`
`dx` `= (du)/2`
`text(When)` `\ \ x = 2,\ ` `u = 3`
  `\ \ x = 1,\ ` `u = 1`

 

`:. int_1^2 x/((2x − 1)^2) \ dx`

`= int_1^3 1/2(u + 1) · 1/(u^2) · (du)/2`

`= 1/4int_1^3 ((u + 1)/(u^2)) du`

`= 1/4 int_1^3 1/u + u^(−2) du`

`= 1/4 [ln u − u^(−1)]_1^3`

`= 1/4 [(ln 3 − 1/3) − (ln 1 − 1)]`

`= 1/4 (ln 3 − 1/3 + 1)`

`= 1/4(ln 3 + 2/3)`

Calculus, EXT1 C2 2008 HSC 2a

Use the substitution  `u = log_e x`  to evaluate  `int_e^(e^2) 1/(x (log_e x)^2)\ dx`.   (3 marks)

Show Answers Only

`1/2`

Show Worked Solution
`u` `= log_e x`
`(du)/(dx)` `= 1/x`
`:. du` `= 1/x\ dx`
`text(When)\ \ \ ` `x = e^2,\ \ ` `u = log_e e^2 = 2`
  `x = e,` `u = 1`

 
`:. int_e^(e^2) 1/(x (log_e x)^2)\ dx`

`= int_1^2 1/(u^2)\ du`

WARNING: Most errors were made in the last stage of substitution in this question. Be careful!

`= int_1^2 u^(-2)\ du`

`= [-1/u]_1^2`

`= [(-1/2) – (-1)]`

`= 1/2`

Calculus, EXT1 C2 2014 HSC 11d

Evaluate  `int_2^5 x/(sqrt(x - 1))\ dx`  using the substitution  `x = u^2 + 1`.   (3 marks) 

Show Answers Only

`20/3`

Show Worked Solution
`x` `= u^2 + 1`
`u^2` `= x – 1`
`u` `= sqrt (x – 1)`
`(du)/(dx)` `= 1/2 (x – 1)^(-1/2)`
  `= 1/(2 sqrt(x – 1))`
`\ \ =>2du` `= dx/sqrt(x – 1)`

 

`text(When)\ \ \ x = 5,` `\ \ u = 2`
`x = 2,` `\ \ u = 1`

`:.\ int_2^5 x/(sqrt(x – 1))\ dx`

`= 2 int_1^2 u^2 + 1\ du`

`= 2 [ (u^3)/3 + u]_1^2`

`= 2 [(8/3 + 2) – (1/3 + 1)]`

`= 20/3`

Calculus, EXT1 C2 2009 HSC 1f

Using the substitution  `u = x^3 + 1`, or otherwise, evaluate  `int_0^2 x^2 e^(x^3 + 1)\ dx`.   (3 marks)

Show Answers Only

`1/3 (e^9\ – e)`

Show Worked Solution
`u` `= x^3 + 1`
`(du)/(dx)` `= 3x^2`
`du` `= 3x^2\  dx`
`text(If)\ \ \ ` `x` `= 2,\ ` `u` `= 9`
  `x` `= 0,\ ` `u` `= 1`

 

`:.\ int_0^2 x^2 e^(x^3 + 1)\ dx`

`=1/3 int_0^2 e^(x^3 + 1) * 3x^2\ dx`

`= 1/3 int_1^9 e^u\ du`

`= 1/3 [e^u]_1^9`

`= 1/3 (e^9\ – e)`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution  `u = e^(3x)`  to evaluate  `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`.   (3 marks)

Show Answers Only

`1/3 (tan^(-1)e\ – pi/4)`

Show Worked Solution
 MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`text(Let)\ \ u = e^(3x)`

`(du)/(dx)` `= 3e^(3x)`
`:.dx` `= (du)/(3e^(3x))`

 

`text(When)` `\ x = 1/3,` `\ u = e^(3 xx 1/3) = e`
  `\ x = 0,` `\ u = e^0 = 1`

 
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`

`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`

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

`= 1/3 [tan^(-1)u]_1^e`

`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`

`= 1/3 (tan^(-1)e\ – pi/4)`

Calculus, EXT1 C2 2013 HSC 5 MC

Which integral is obtained when the substitution  `u = 1 + 2x`  is applied to  `int x sqrt(1 + 2x)\ dx`?

  1. `1/4 int (u - 1) sqrt u\ du`
  2. `1/2 int (u - 1) sqrt u\ du` 
  3. `int (u - 1) sqrt u\ du`
  4. `2 int (u - 1) sqrt u\ du`
Show Answers Only

`A`

Show Worked Solution
`text(Let)\ \ u` `= 1 + 2x`
`:.x` `= 1/2 (u – 1)`
`(du)/(dx)` `= 2`
`:.dx` `= 1/2\ du`

 
`int x sqrt(1 + 2x)\ dx`

`=int 1/2 (u – 1) xx u^(1/2) xx 1/2\ du`

`= 1/4 int (u – 1) sqrt u\ du`
 

`=>  A`

Calculus, EXT1 C2 2010 HSC 1e

Use the substitution  `u = 1 - x`  to evaluate  `int_0^1 x sqrt(1 - x)\ dx`.   (3 marks)

Show Answers Only

`4/15`

Show Worked Solution
`u = 1 – x` `\ \ \ \ \ => x = 1 – u`
`(du)/(dx) = -1` `\ \ \ \ \ => du = – dx`

 

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

 
`:. int_0^1 x sqrt(1 – x)\ dx`

`= – int_1^0 (1 – u) u^(1/2)\ du`

`= int_1^0 (u – 1) u^(1/2)\ du`

`= int_1^0 (u^(3/2) – u^(1/2))\ du`

`= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^0`

`= [0 – (2/5 – 2/3)]`

`= – (6/15 – 10/15)`

`= 4/15`

Calculus, EXT1 C2 2011 HSC 1d

Using the substitution  `u = sqrtx`, evaluate  `int_1^4 (e^(sqrtx))/(sqrtx)\ dx`.   (3 marks) 

Show Answers Only

`2e (e – 1)`

Show Worked Solution
`u` `= sqrtx = x^(1/2)`
`(du)/(dx)` `= 1/2 x^(-1/2) = 1/(2 sqrtx)`
 `du` `= (dx)/(2sqrtx)`
`:.2du` `= (dx)/(sqrtx)`

 

`text(When)\ \ \ ` `x=4,\ \ ` `u = 2`
  `x = 1,` `\ x = 1`

 

`:. int_1^4 (e^(sqrtx))/(sqrtx)\ dx`
`= int_1^2 e^u xx 2\ du`
`= 2 [e^u]_1^2`
`= 2 [e^2 – e^1]`
`= 2e (e – 1)`

Calculus, EXT1 C2 2012 HSC 11d

Use the substitution  `u = 2 - x`  to evaluate  `int_1^2 x (2 - x)^5\ dx`.   (3 marks)  

Show Answers Only

 `4/21`

Show Worked Solution
`u` `=2-x`
`:. x` `= 2 – u`
`(du)/dx` `= -1`
`:.dx` `=-du`

 

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

 

`:. int_1^2 x (2 – x)^5\ dx` `= int_1^0 – (2 – u) u^5\ du`
  `= int_1^0 u^6 – 2u^5\ du`
  `= [1/7 u^7 – 2/6 u^6]_1^0`
  `= [0 – (1/7 – 1/3)]`
  `= – (3/21 – 7/21)`
  `= 4/21`