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

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

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

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

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

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

 

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

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

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

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

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

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

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

 

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

Calculus, EXT2 C1 2022 SPEC2 7 MC

Using the substitution  `u=1+e^x, \int_0^{\log _e 2} \frac{1}{1+e^x}dx`  can be expressed as

  1. `\int_0^{\log _e 2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
  2. `\int_2^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  3. `\int_1^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  4. `\int_2^3\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
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`D`

Show Worked Solution

`u=1+e^x\ \ =>\ \ \frac{du}{dx}=e^x\ \ =>\ \ dx=\frac{du}{e^x}=\frac{du}{u-1}`

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

`text{When}\ \ x=\log_e 2,\ \ u=1+e^{\log_e 2} = 3`

`\int_0^{\log _e 2} \frac{1}{1+e^x}dx = \int_2^3\left(\frac{1}{u}*\frac{1}{u-1}\right) du= \int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`

`=>D`

Calculus, EXT2 C1 2019 HSC 15a

  1. Show that
     
    `qquad int_(-a)^a (f(x))/(f(x) + f(-x))\ dx = int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`.  (2 marks)

  2. Hence, or otherwise, evaluate
     
    `qquad int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`.  (2 marks)

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

i.    `text(Let)\ \ u = -x \ => \ du = -dx`

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

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

`int_(-a)^a (f(x))/(f(x) + f(-x))\ dx` `= -int_a^(-a) (f(-u))/(f(-u) + f(u))\ du`
  `= int_(-a)^a (f(-u))/(f(u) + f(-u))\ du`
  `= int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`

 

ii.    `text(Let)\ \ I = int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`

`2I` `= int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx + int_(-1)^1 (e^(-x))/(e^x + e^(-x))\ dx`
`2I` `= int_(-1)^1 (e^x + e^(-x))/(e^x + e^(-x))\ dx`
`2I` `= [x]_(-1)^1`
`2I` `= 2`

 
`:. int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx = 1`