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Calculus, 2ADV C4 2024 HSC 14

The curves  `y=(x-1)^2`  and  `y=5-x^2`  intersect at two points, as shown in the diagram.
 

  1. Find the `x`-coordinates of the points of intersection of the two curves.   (1 mark)

  2. Find the area enclosed by the two curves.   (3 marks)

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a.   `x=2\ \text{and}\ -1`

b.   `9\ \text{units}^2`

Show Worked Solution

a.   `y=(x-1)^2, \ y=5-x^2`

\(\text{Intersection occurs when:}\)

`(x-1)^2` `=5-x^2`  
`x^2-2x+1` `=5-x^2`  
`2x^2-2x-4` `=0`  
`2(x-2)(x+1)` `=0`  

 
`x=2\ \text{and}\ -1`
 

b.    `\text{Area}` `= \int_{-1}^{2} (5-x^2)-(x-1)^2\ dx` 
    `=\int_{-1}^{2} 5-x^2-x^2+2x-1\ dx`
    `=\int_{-1}^{2}4-2x^2+2x\ dx`
    `=[4x-\frac{2}{3}x^3+x^2]_{-1}^{2}`
    `=[(8-\frac{16}{3}+4)-(-4+\frac{2}{3}+1)]`
    `=\frac{20}{3}-(-\frac{7}{3})`
    `=9\ \text{u}^2`

Calculus, 2ADV C4 2022 HSC 16

The parabola `y=x^2` meets the line `y=2 x+3` at the points `(-1,1)` and `(3,9)` as shown in the diagram.
 

Find the area enclosed by the parabola and the line.  (3 marks)

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`32/3\ text{u}^2`

Show Worked Solution
`A` `=int_(-1)^(3) 2x+3-x^2\ dx`  
  `=[x^2+3x-1/3 x^3]_(-1)^3`  
  `=(9+9-27/3)-(1-3+1/3)`  
  `=9-(-5/3)`  
  `=32/3\ text{u}^2`  

Calculus, 2ADV C4 2020 HSC 30

The diagram shows two parabolas  `y = 4x - x^2`  and  `y = ax^2`, where  `a > 0`. The two parabolas intersect at the origin, `O`, and at `A`.
 


 

  1. Show that the `x`-coordinate of  `A`  is  `4/(a + 1)`.  (2 marks)

  2. Find the value of  `a`  such that the shaded area is `16/3`.  (4 marks)

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  1. `text(See Worked Solutions)`
  2. `sqrt2 – 1`
Show Worked Solution

a.   `text(Intersection occurs when)`

`4x – x^2` `= ax^2`
`x^2(a + 1) – 4x` `= 0`
`x[x(a + 1) – 4]` `= 0`
`x(a+1)-4` `=0\ \ \ text(or)`   `x=0`
`:. x_A` `=4/(a + 1)`  

♦ Mean mark part (b) 48%.

 

b.    `text(Area)` `= int_0^(4/(a + 1)) 4x – x^2\ dx – int_0^(4/(a + 1)) ax^2\ dx`
  `16/3` `= int_0^(4/(a + 1)) 4x – (1 + a)x^2\ dx`
  `16/3` `= [2x^2 – ((1 + a)/3) x^3]_0^(4/(a + 1))`
  `16/3` `= 2(4/(a + 1))^2 – ((1 + a)/3)(4/(a + 1))^3`
  `16/3` `= 32/((a + 1)^2) – 64/3 · 1/((a + 1)^2)`
  `16` `= 96/((a + 1)^2) – 64/((a + 1)^2)`
`16(a + 1)^2` `= 32`
`(a + 1)^2` `= 2`
`a + 1` `= sqrt2,\ \ \ (a > 0)`
`:. a` `= sqrt2 – 1`

Calculus, 2ADV C4 2017 HSC 14d

The shaded region shown is enclosed by two parabolas, each with `x`-intercepts at  `x = –1`  and  `x = 1`.

The parabolas have equations  `y = 2k (x^2 - 1)`  and  `y = k(1 - x^2)`, where  `k > 0`.
  


 

Given that the area of the shaded region is 8, find the value of `k`.  (3 marks)

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`k = 2`

Show Worked Solution
`text(Area)` `= int_(-1)^1 k (1 – x^2) dx – int_(-1)^1 2k (x^2 – 1) dx`
`8` `= 2 int_0^1 k – kx^2 – 2kx^2 + 2k\ dx`
`8` `= 2 int_0^1 3k – 3kx^2\ dx`
`8` `= 2 [3kx – kx^3]_0^1`
`8` `= 2 [(3k – k) – 0]`
`8` `= 4k`
`:. k` `= 2`

Calculus, 2ADV C4 SM-Bank 1 MC

The diagram below is the graph of  `y = x^2 - x - 6`
 

Integration, 2UA SM-Bank 01
 

What is the correct expression for the area bounded by the `x`-axis and the graph  `y = x^2 - x - 6`  between  `0 <= x <= 4`?

  1. `A = int_0^4 x^2 - x - 6\ dx`
  2. `A = |\ int_0^3 x^2 - x - 6\ dx\ | + int_3^4 x^2 - x - 6\ dx`
  3. `A = int_0^3 x^2 - x - 6\ dx  + |\ int_3^4 x^2 - x - 6\ dx|`
  4. `A = |\ int_0^4 x^2 - x - 6\ dx\ |`
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`=> B`

Show Worked Solution
`y` `= x^2 – x – 6`
  `= (x – 3)(x + 2)`

 

`text(Graph intersects the)\ xtext(-axis at)\ (3,0)`

`:. A = |\ int_0^3 x^2 – x – 6\ dx\ | + int_3^4 x^2 – x – 6\ dx`

`=> B`

Calculus, 2ADV C4 2015 HSC 16a

The diagram shows the curve with equation  `y = x^2-7x + 10`. The curve intersects the `x`-axis at points `A and B`. The point `C` on the curve has the same `y`-coordinate as the `y`-intercept of the curve.
 


 

  1. Find the `x`-coordinates of points `A and B.`  (1 mark)

  2. Write down the coordinates of `C.`  (1 mark)

  3. Evaluate  `int _0^2 (x^2-7x + 10)\ dx.`  (1 mark)

  4. Hence, or otherwise, find the area of the shaded region.  (2 marks)

Show Answers Only
  1. `A = 2,\ \ B = 5`
  2. `(7, 10)`
  3. `8 2/3`
  4. `16 1/3\ text(u²)`
Show Worked Solution
i.    `y` `= x^2-7x + 10`
  `= (x-2) (x-5)`

 
`:.x = 2 or 5`

`:.\ \ x text(-coordinate of)\ \ A = 2`

`x text(-coordinate of)\ \ B = 5`

 

ii.    `y\ text(intercept occurs when)\ \ x = 0`

`=>y text(-intercept) = 10`
 

`C\ text(occurs at intercept:)`

`y` `= x^2-7x + 10` `\ \ \ \ \ text{…  (1)}`
`y` `= 10` `\ \ \ \ \ text{…  (2)}`

 
`(1) = (2)`

`x^2-7x + 10` `= 10`
`x^2-7x` `= 10`
`x (x-7)` `= 10`

 
`x = 0 or 7`

`:.\ C\ \ text(is)\ \ (7, 10)`

 

iii.  `int_0^2 (x^2 – 7x + 10)\ dx`

`= [1/3 x^3 – 7/2 x^2 + 10x]_0^2`

`= [(1/3 xx 2^3 – 7/2 xx 2^2 + 10 xx 2) – 0]`

`= 8/3 – 14 + 20`

`= 8 2/3`

 

iv.  

`A_1 = A_2`

♦ Mean mark 49%.

`A_2 = 8 2/3\ text(u²)\ \ \ \ text{(from part (iii))}`

`text(Let)\ \ D\ \ text(be)\ \ (7, 0)`

`text(Shaded Area)`

`= text(Area of)\ \ Delta ACD – A_2`

`= 1/2 bh – 8 2/3`

`= 1/2 xx 5 xx 10 – 8 2/3`

`= 16 1/3\ text(u²)`

Calculus, 2ADV C4 2015 HSC 7 MC

The diagram shows the parabola  `y = 4x - x^2`  meeting the line  `y = 2x`  at  `(0, 0)`  and  `(2, 4)`.
 

2015 2ua 7 mc
 

Which expression gives the area of the shaded region bounded by the parabola and the line?

  1. `int_0^2 x^2 - 2x\ dx`
  2. `int_0^2 2x - x^2\ dx`
  3. `int_0^4 x^2 - 2x\ dx`
  4. `int_0^4 2x - x^2\ dx`
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`B`

Show Worked Solution
`text(Shaded Area)` `= int_0^2 4x – x^2\ dx – int_0^2 2x\ dx`
  `= int_0^2 4x – x^2 – 2x\ dx`
  `= int_0^2 2x – x^2\ dx`

 

`=> B`

Calculus, 2ADV C4 2005 HSC 8b

2005 8b

The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola  `y = x^2– 3x + 2`, and the `x`-axis.

By considering the difference of two areas, find the area of the shaded region.  (3 marks)

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`(pi − 5/6)\ \ text(u²)`

Show Worked Solution

`text(Shaded Area = Area in the quarter circle less)`

`text(the area below the parabola between)\ x= 0 and 1.`

`text(Area of)\ 1/4\ text(circle)` `= 1/4 xx pir^2`
  `= 1/4 xx pi xx 2^2`
  `= pi\ \ \ text(u²)`

 

`text(Area below the parabola between)\ x= 0 and 1`

`=int_0^1y\ dx`

`= int_0^1x^2 − 3x + 2\ dx`

`= [x^3/3 − 3/2x^2 + 2x]_0^1`

`= [(1/3 − 3/2 + 2) − 0]`

`= 5/6`

 

`:.\ text(Shaded Area) = (pi − 5/6)\ \ \ text(u²)`

Calculus, 2ADV C4 2014 HSC 12d

The parabola  `y = −2x^2 + 8x`  and the line  `y = 2x`  intersect at the origin and at the point  `A`.
  


  

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

  2. Calculate the area enclosed by the parabola and the line.     (3 marks)

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  1. `x=3`
  2. `9\ text(u²)`
Show Worked Solution
i.  

`text(Need to find)\ x\ text(-co-ord of)\ A`

`y` `= 2x\ \ \ \ \ …\ text{(i)}`
`y` `= -2x^2 + 8x\ \ \ \ \ …\ text{(ii)}`

 
`text(Subst)\ y = 2x\ text(from)\ text{(i)}\ text(into)\ text{(ii)}`

`-2x^2 + 8x` `= 2x`
`-2x^2 + 6x` `= 0`
`-2x (x\ – 3)` `= 0`

  
`:.\ x = 0\ text(or)\ 3`

`:.\ x\ text(-coordinate of)\ A\ text(is 3)`

 

ii. `text(Area)` `= int_0^3 (-2x^2 + 8x)\ dx\ – int_0^3 2x\ dx`
    `= int_0^3 (-2x^2 + 6x)\ dx`
    `= [-2/3x^3 + 3x^2]_0^3`
    `= [(-2/3(3^3) + 3(3^2))\ – (0 + 0)]`
    `= -18 + 27`
    `= 9\ text(u²)`

Calculus, 2ADV C4 2012 HSC 13b

The diagram shows the parabolas `y = 5x - x^2` and `y = x^2 - 3x`. The parabolas intersect at

the origin `O` and the point `A`. The region between the two parabolas is shaded. 
 

2012 13b
 

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

  2. Find the area of the shaded region.   (3 marks)

Show Answers Only
  1. `4`
  2. `64/3 \ text(u²)`
Show Worked Solution

i.     `y = x^2 -3x\ \ …\ (1)`

`y = 5x – x^2\ \ …\ (2)`

`text(Solve:)\ \ (1)=(2)`

`x^2 – 3x` `= 5x  – x^2`
`2x^2 – 8x` `= 0`
`2x(x – 4)` `=0`
`x` `= 0  \ text(or) \ 4`
MARKER’S COMMENT: Less errors were made in part (ii) by students who simplified the definite integral before integrating, as shown in the worked solution. 

 
`:. x text(-coordinate of)  \ A  \ text(is) \  4`

 

ii.   `text(Area)` `= int_0^4 (5x – x^2) dx  –  int_0^4 (x^2 – 3x) dx`
  `= int_0^4 (5x – x^2 – x^2 + 3x) dx`
  `= int_0^4 (8x – 2x^2) dx`
  `= [4x^2 – 2/3x^3]_0^4`
  `= [(4 xx 4^2) – (2/3 xx 4^3)]`
  `= [ 64\ – 128/3 ]`
  `= 64/3 \ text(u²)`