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Calculus, 2ADV C4 EQ-Bank 25

  1. Differentiate  \(y=x^2\, \log _e x\).   (2 marks)

  2. Hence, or otherwise, find \(\displaystyle \int_1^e x\, \log _e x\, d x\).   (2  marks)

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a.    \(\dfrac{d y}{d x} =x+2 x\, \log _e x\)

b.    \(\dfrac{e^2}{4}+\dfrac{1}{4}\)

Show Worked Solution

a.    \(y=x^2\, \log _e x\)

\(\dfrac{d y}{d x}\) \(=x^2 \cdot \dfrac{1}{x}+2 x\, \log _e x\)
  \(=x+2 x\, \log _e x\)

 
b.    \(\text{Using part a.}\)

\(\displaystyle \int x+2 x\, \log _e x\, d x\) \(=x^2\, \log _e x+c\)  
\(\displaystyle \int x\, d x+2 \int x\, \log _e x\, d x\) \(=x^2\, \log _e x+c\)  
\(\displaystyle 2 \int_1^e x\, \log _e x\, d x\) \(=\left[x^2\, \log _e x\right]_1^e-\displaystyle \int_1^e x\, d x\)
\(\displaystyle \int_1^e x\, \log _e x\, d x\) \(=\dfrac{1}{2}\left[\left(e^2 \cdot 1\right)-0\right]-\dfrac{1}{2}\left[\dfrac{x^2}{2}\right]_1^e\)
  \(=\dfrac{e^2}{2}-\left(\dfrac{e^2}{4}-\dfrac{1}{4}\right)\)
  \(=\dfrac{e^2}{4}+\dfrac{1}{4}\)

Calculus, 2ADV C4 EQ-Bank 1

  1. Differentiate  \(y=x e^x\).   (2 marks)

  2. Hence, or otherwise, find \(\displaystyle \int_1^e x e^x d x\).   (2 marks)
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a.  \(\dfrac{d y}{d x}=e^x+x e^x\)

b.  \(e^e(e-1)\)

Show Worked Solution

a.    \(y=x e^x\)

\(\dfrac{d y}{d x}=e^x+x e^x\)
 

b.    \(\text{Using part a.}\)

\(\displaystyle\int e^x+x e^x\, d x\) \(=x e^x+c\)
\(\displaystyle\int_1^e e^x\, d x+\int_1^e x e^x\, d x\) \(=\left[x e^x\right]_1^e\)
\(\displaystyle\int_1^e x e^x\, d x\) \(=\left[x e^x\right]_1^e-\left[e^x\right]_1^e\)
  \(=\left[e \cdot e^e-e\right]-\left[e^e-e\right]\)
  \(=e \cdot e^e-e^e\)
  \(=e^e(e-1)\)

Calculus, 2ADV C4 2020 HSC 18

  1. Differentiate  `e^(2x) (2x + 1)`.   (2 marks)

  2. Hence, find  `int(x + 1)e^(2x)\ dx`.   (1 marks)

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a.    `4e^(2x)(x + 1)`

b.    `1/4 e^(2x)(2x + 1) + c`

Show Worked Solution
a.    `y` `= e^(2x) (2x + 1)`
  `(dy)/(dx)` `= 2e^(2x)(2x + 1) + 2e^(2x)`
    `= 2e^(2x)(2x + 2)`
    `= 4e^(2x)(x + 1)`

♦ Mean mark part (b) 40%.

 

b.     `int(x + 1)e^(2x)dx` `= 1/4 int 4e^(2x)(x + 1)`
    `= 1/4 e^(2x)(2x + 1) + c`

Calculus, 2ADV C4 2019 HSC 13c

  1.  Differentiate `(ln x)^2`.   (2 marks)

  2.  Hence, or otherwise, find `int(ln x)/x\ dx`.   (1 mark)

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i.    `(2 ln x)/x`

ii.   `1/2 (ln x)^2 + C`

Show Worked Solution

i.    `y= (ln x)^2`

`(dy)/(dx)= 2 xx 1/x xx ln x= (2 ln x)/x`

♦ Mean mark (ii) 49%.

 

ii.   `int (ln x)/x\ dx=1/2 int (2 ln x)/x dx= 1/2 (ln x)^2 +C`

Calculus, 2ADV C4 2016 HSC 12d

  1. Differentiate  `y = xe^(3x)`.   (1 mark)

  2. Hence find the exact value of  `int_0^2 e^(3x) (3 + 9x)\ dx`.   (2 marks)

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a.    `e^(3x) (1 + 3x)`

b.    `6e^6`

Show Worked Solution

a.    `y = xe^(3x)`

`text(Using product rule:)`

`(dy)/(dx)= x · 3e^(3x) + 1 · e^(3x)= e^(3x) (1 + 3x)`
 

b.    `int_0^2 e^(3x) (3 + 9x)\ dx`

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

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6-0)= 6e^6`