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Calculus, SPEC1 2023 VCAA 5

Evaluate  \(
\begin{aligned}
\int_1^2x^2\, \log_e(x)\ dx. \\
\end{aligned}
\)   (3 marks)

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\(\dfrac{8}{3}\log_e(2)-\dfrac{7}{3} \)

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  \(u\) \(= \log_e x,\) \(\ \ \ \ u^{′}\) \(= \dfrac{1}{x}\)
  \(v^{′}\) \(= x^2,\) \(v\) \(= \dfrac{1}{3} x^3\)
\[\int_1^2 x^2\,\log_e (x)  \,dx\] \[= uv-\int u^{′} v \ dx\]
  \[= \Big{[}\dfrac{x^3}{3}\,\log_e x\Big{]}_1 ^2-\dfrac{1}{3} \int_1 ^2 \dfrac{1}{x} \times x^3\ dx\]
  \[= \Big{[} \dfrac{8}{3} \log_e 2-0\Big{]}-\dfrac{1}{9} \Big{[}x^3\Big{]}_1^2\]
  \(=\dfrac{8}{3} \log_e 2-\dfrac{1}{6}(8-1) \)
  \(=\dfrac{8}{3} \log_e 2-\dfrac{7}{9} \)

Calculus, EXT2 C1 2020 HSC 11b

Use integration by parts to evaluate  `int_1^e x ln x \ dx`.   (3 marks)

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`frac{e^2 + 1}{4}`

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`u = ln \ x` `v′ = x`
`u′ = frac{1}{x}` `v = frac{x^2}{2}`

 

`int _1^e x \ ln \ x \ dx` `= [ frac{x^2}{2} · ln \ x ]_1^e – int_1^e frac{x^2}{2} · frac{1}{x}\ dx`
  `= [frac{e^2}{2} ln \ e – frac{1}{2} ln 1]- int_1^e frac{x}{2}\ dx`
  `= frac{e^2}{2} – [ frac{x^2}{4}]_1^e`
  `= frac{e^2}{2} – ( frac{e^2}{4} – frac{1}{4} )`
  `= frac{e^2 + 1}{4}`

Calculus, EXT2 C1 2005 HSC 1c

Use integration by parts to evaluate  `int_1^e x^7 log_e x  dx`.   (3 marks)

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`(7e^8 – 1)/(64)`

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  `u` `= log_e x,` `\ \ \ \ u^(′)` `= (1)/(x)`
  `v^(′)` `= x^7,` `v` `= (1)/(8) x^8`

 

`int_1^e x^7\ log_e x  dx` `= uv-int u^(′) v \ dx`
  `= [(x^8)/(8) ⋅ log_e x]_1 ^e-(1)/(8) int_1 ^e (1)/(x) ⋅ x^8 dx`
  `= ((e^8)/(8) log_e e-(1)/(8) log_e 1)-(1)/(8)[(1)/(8) x^8]_1 ^e`
  `= (e^8)/(8)-(1)/(8) ((e^8)/(8)-(1)/(8))`
  `= (e^8)/(8)-((e^8-1)/(64))`
  `= (7e^8 +1)/(64)`

Calculus, EXT2 C1 2003 HSC 1b

Use integration by parts to find   `int x^3 log_e x  dx`   (3 marks)

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`(x^4 log_e x)/(4)-(x^4)/(16) + C`

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`u` `= log_e x,` `\ \ \ \ u^{′}` `= (1)/(x)`
`v^{′}` `= x^3,` `v` `= (x^4)/(4)`

 

`int x^3 log_e x  dx` `= uv-int u^{′}v \ dx`
  `= (x^4)/(4)  log_e x-int (1)/(x) ⋅ (x^4)/(4)  dx`
  `= (x^4 log_e x)/(4)-(1)/(4) int x^3 dx`
  `= (x^4 log_e x)/(4)-(x^4)/(16) + C`

Calculus, EXT2 C1 2014 HSC 16c

Find  `int(ln\ x)/((1 + ln\ x)^2)\ dx`.  (3 marks)

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`x/(1+ln x) +c`

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`I=int(ln\ x)/((1 + ln\ x)^2)\ dx`

♦♦ Mean mark 31%.
STRATEGY: This is as challenging as integration gets … a substitution followed by applying integration by parts. Note that simply substituting `u=ln x` gets you a full mark.
`text(Let)\ \ \ u=` `ln x`
`(du)/(dx)=` `1/x=1/e^u`
`du=` `1/e^u\ dx`
`dx=` `e^u\ du`
`I` `=int (u e^u)/(1+u)^2\ du`
  `=int ((1+u) e^u)/(1+u)^2\ du- int e^u/(1+u)^2\ du`
  `=int e^u/(1+u)\ du -int e^u/(1+u)^2\ du`
 

`text{Using integration by parts:}`

`u` `=-e^u`    `u′` `=-e^u`
`v′` `=-(1+u)^-2` `v` `=(1+u)^-1`
 `:.I` `=int e^u/(1+u)\ du – ((-e^u)/(1+u) + int (e^u)/(1+u)\ du)`
  `=e^u/(1+u) +c`
  `=x/(1+ln x) +c`