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Calculus, 2ADV C4 2018* HSC 15c

The shaded region is enclosed by the curve  `y = x^3 - 7x`  and the line  `y = 2x`, as shown in the diagram. The line  `y = 2x`  meets the curve  `y = x^3 - 7x`  at  `O(0, 0)`  and  `A(3, 6)`. Do NOT prove this.
 


 

  1.  Use integration to find the area of the shaded region.  (2 marks)

  2. Use the Trapezoidal rule and four function values to approximate the area of the shaded region.  (2 marks)

The point `P` is chosen on the curve  `y = x^3 − 7x`  so that the tangent at `P` is parallel to the line  `y = 2x`  and the `x`-coordinate of `P` is positive

  1.  Show that the coordinates of `P` are  `(sqrt 3, -4 sqrt 3)`.  (2 marks)

  2.  Using the perpendicular distance formula  `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`,  find the area of  `Delta OAP`.  (2 marks)

Show Answers Only
  1. `81/4\ text(units²)`
  2. `18\ text(u²)`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `9 sqrt 3\ text(units²)`
Show Worked Solution
i.   `text(Area)` `= int_0^3 2x – (x^3 – 7x)\ dx`
    `= int_0^3 9x – x^3\ dx`
    `= [9/2 x^2 – 1/4 x^4]_0^3`
    `= [(9/2 xx 3^2 – 1/4 xx 3^4) – 0]`
    `= 81/2 – 81/4`
    `= 81/4\ text(units²)`

 

ii.  `f(x) = 9x – x^3`

`text(Area)` `~~ 1/2[0 + 2(8 + 10) + 0]`
  `~~ 1/2(36)`
  `~~ 18\ text(u²)`

 

iii.   `y = x^3 – 7x`

`(dy)/(dx) = 3x^2 – 7`

`text(Find)\ \ x\ \ text(such that)\ \ (dy)/(dx) = 2`

`3x^2 – 7` `= 2`
`3x^2` `= 9`
`x^2` `= 3`
`x` `= sqrt 3 qquad (x > 0)`

 

`y` `= (sqrt 3)^3 – 7 sqrt 3`
  `= 3 sqrt 3 – 7 sqrt 3`
  `= -4 sqrt 3`

 
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

 

iv.  

 

`text(dist)\ OA` `= sqrt((3 – 0)^2 + (6 – 0)^2)`
  `= sqrt 45`
  `= 3 sqrt 5`

 
`text(Find)\ _|_\ text(distance of)\ \ P\ \ text(from)\ \ OA`

`P(sqrt 3, -4 sqrt 3),\ \ 2x – y=0`

`_|_\ text(dist)` `= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|`
  `= (6 sqrt 3)/sqrt 5`
   
`:.\ text(Area)` `= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5`
  `= 9 sqrt 3\ text(units²)`

Calculus, 2ADV C4 2013 HSC 13b

The diagram shows the graphs of the functions  `f(x) = 4x^3-4x^2 +3x`  and  `g(x) = 2x`. The graphs meet at  `O`  and at  `T`.
 

2013 13b

  1. Find the  `x`-coordinate of  `T`.   (1 mark)

  2. Find the area of the shaded region between the graphs of the functions  `f(x)`  and  `g(x)`.   (3 marks)

Show Answers Only
  1. `T\ text(is)\ (1/2, 1)`
  2. `text(Shaded area between the curves is)\ 1/48\ text(u²)`
Show Worked Solution
i. `T\ text(occurs when)\ \ f(x) = g(x)`
`4x^3-4x^2 + 3x` `=2x`
`4x^3-4x^2 + x` `=0`
`x(4x^2-4x + 1)` `=0`
`x(2x-1)^2` `=0`

 

`2x-1=0 \ \ => \ x=1/2`

`text(Substitute)\ \ x=1/2\ \ text(into)\ g(x)`

`g(1/2) = 2 xx 1/2 = 1`

`:.\ T (1/2, 1)`

 

ii. `text(Shaded Area)` `= int_0^(1/2) (f(x)-g(x)) \ dx`
    `= int_0^(1/2) (4x^3-4x^2 + x) \ dx`
    `= [x^4-4/3x^3 + 1/2x^2]_0^(1/2)`
    `= [((1/2)^4-4/3(1/2)^3 + 1/2(1/2)^2)-0]`
    `= 1/16-1/6 + 1/8`
    `= 1/48\ text(u²)`

 
`:.\ text(Shaded area between the curves is)\ 1/48\ text(u²)`