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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)

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

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

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

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 2015 HSC 11e

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

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

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`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)

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`1/2`

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`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 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)

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`1/3 (e^9\ – e)`

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`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)

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`1/3 (tan^(-1)e\ – pi/4)`

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 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 2011 HSC 1d

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

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`2e (e – 1)`

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`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)`