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Calculus, 2ADV C4 2023 MET2 6 MC

Suppose that \(\displaystyle \int_{3}^{10} f(x)\,dx=C\)  and  \(\displaystyle \int_{7}^{10} f(x)\,dx=D\). The value of \(\displaystyle \int_{7}^{3} f(x)\,dx\) is

  1. \(C+D\)
  2. \(C+D-3\)
  3. \(C-D\)
  4. \(D-C\)
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\(D\)

Show Worked Solution

\(\text{Given }\displaystyle \int_{3}^{10} f(x)\,dx=C\ \ \text{and}\ \displaystyle \int_{7}^{10} f(x)\,dx=D\)

\(\text{We can deduce:}\)

 \(\displaystyle \int_{3}^{10} f(x)\,dx\) \(=\displaystyle \int_{3}^{7} f(x)\,dx+\displaystyle \int_{7}^{10} f(x)\,dx\)
\(C\) \(=\displaystyle \int_{3}^{7} f(x)\,dx+D\)
\(C-D\) \(=\displaystyle \int_{3}^{7} f(x)\,dx\)
\(\therefore\ \displaystyle \int_{7}^{3} f(x)\,dx\) \(=D-C\)

 
\(\Rightarrow D\)


♦ Mean mark 49%.
MARKER’S COMMENT: 41% of students chose option C incorrectly assuming \(\int_{7}^3 f(x)\,dx=\int_{3}^7 f(x)\,dx\).

Calculus, 2ADV C4 2023 HSC 5 MC

The diagram shows the graph `y=f(x)`, where `f(x)` is an odd function.

The shaded area is 1 square unit.

The number `a`, where `a > 1`, is chosen so that `int_0^a f(x)\ dx=0`.
 

       

What is the value of `int_{-a}^1 f(x)\ dx` ?

  1. `-1`
  2. `0`
  3. `1`
  4. `3`
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`A`

Show Worked Solution

`text{Since}\ \ int_0^a f(x)\ dx=0\ and \ int_0^1 f(x)\ dx=-1\ \ text{(given)}`

`int_1^a f(x)\ dx=1`

`:. int_{-a}^{-1} f(x)\ dx=-1\ \ text{(f(x) is odd)}`

`:. int_{-a}^1 f(x)\ dx=-1`

`=>A`

♦ Mean mark 49%.

Calculus, 2ADV C4 2022 HSC 8 MC

The graph of the even function  `y=f(x)`  is shown.

The area of the shaded region `A` is `1/2` and the area of the shaded region `B` is `3/2`.

What is the value of `int_(-2)^(2)f(x)\ dx`?

  1. `4`
  2. `2`
  3. `–2`
  4. `–4`
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`C`

Show Worked Solution

`text{Areas under the}\ x text{-axis are negative}`

`int_0 ^2 f(x)\ dx = 1/2-3/2=-1`
 

`text{S}text{ince}\ \ f(x)\ \ text{is even:}`

`int_(-2) ^2 f(x)\ dx = 2int_0 ^2 f(x)\ dx = -2`

`=>C`


Mean mark 52%.

Calculus, 2ADV C4 2020 HSC 7 MC

The diagram show the graph  `y = f(x)`, which is made up of line segments and a semicircle.
 

What is the value of  `int_0^12 f(x)\ dx`?

  1.  `24 + 2pi`
  2.  `24 + 4pi`
  3.  `30 + 2pi`
  4.  `30 + 4pi`
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`A`

Show Worked Solution

`text(Consider the interval between)\ \ x=8 and x=12:`

`text(Area above and below the)\ xtext(-axis are equal.)`

`int_8^12 f(x) = 0`

♦ Mean mark 48%.
 

`:. int_0^12 f(x)\ dx` `= int_0^8 f(x)\ dx`
  `=\ text(Area of rectangle + area of semi-circle)`
  `= 8 xx 3 + 1/2 xx pi xx 2^2`
  `= 24 + 2pi`

 
`=>A`

Calculus, 2ADV C4 2019 MET-N 15 MC

The area bounded by the graph of  `y = f(x)`, the line  `x = 2`, the line  `x = 8`  and the `x`-axis, as shaded in the diagram below, is  `3log_e(13)`

The value of  `int_4^10 3 f(x - 2)\ dx`  is

  1.  `3log_e(13)`
  2.  `9log_e(13)`
  3.  `6log_e(39)`
  4.  `9log_e(11)`
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`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x – 2) = f(x) \ text(shifted 2 units to right.)`

`:. \ int_2^8 f(x)\ dx = int_4^10 f(x – 2)\ dx`

`:. \ 3 int_4^10 f(x – 2)\ dx` `= 3 xx 3log_e 13`
  `= 9 log_e 13`

 
`=> \ B`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are  `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for  `t >= 6`.
 


 

  1. The object is initially at the origin. During which time(s) is the displacement of the object decreasing?  (1 mark)

  2. If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer.  (2 marks)

  3. Sketch the displacement, `x`, as a function of time.  (2 marks)

Show Answers Only
  1. `t > 5\ \ text(seconds)`
  2. `7.2\ \ text(seconds)`
  3.    
Show Worked Solution

i.  `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

 

ii.  `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Considering displacement from)\ B\ text(to)\ D:`

`text(S)text(ince the area below the graph from)`

`B\ text(to)\ C\ text(equals the area above the)`

`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`

`text(in displacement from)\ B\ text(to)\ D.`

 

`text(Considering)\ t >= 6`

`text(Time required to return to origin)`

`t` `= d/v`
  `= 7/5`
  `= 1.4\ \ text(seconds)`

 

`:.\ text(The particle returns to the origin after 7.4 seconds.)`
   

iii.   

Calculus, 2ADV C4 2018 HSC 7 MC

The diagram shows the graph of  `y = f(x)`  with intercepts at  `x = -1, 0, 3 and 4.`
 

 
The area of shaded region `R_1` is 2.

The area of shaded region `R_2` is 3.

It is given that `int_0^4 f(x)\ dx = 10`.

What is the value of `int_(-1)^3 f(x)\ dx?`

  1. 5
  2. 9
  3. 11
  4. 15
Show Answers Only

`C`

Show Worked Solution

`int_0^4 f(x)\ dx = 10`

♦ Mean mark 36%.

`:. R_3 – R_2` `= 10`
`R_3` `= 13`

 

`int_(-1)^3 f(x)\ dx` `= R_3 – R_1`
  `= 13 – 2`
  `= 11`

 `=>  C`

Calculus, 2ADV C4 2012 HSC 10 MC

The graph of  `y = f(x)`  has been drawn to scale for  `0 <= x <= 8`.
 

2012 10 mc
 

Which of the following integrals has the greatest value? 

  1. `int_0^1 f(x) \ dx` 
  2. `int_0^2 f(x) \ dx`  
  3. `int_0^7 f(x) \ dx`  
  4. `int_0^8 f(x) \ dx`  
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`B`

Show Worked Solution

`text(S)text(ince the integrals measure the net area under)`

`text(the graph and above the)\ x text(-axis)\ text{(i.e. below the}`

`x text{-axis is a negative value.)}`

`=>  B`

Calculus, 2ADV C4 2013 HSC 14d

The diagram shows the graph  `f(x)`.
 

2013 14d
 

What is the value of  `a`, where  `a > 0`, so that  `int_-a^a f(x)\ dx = 0`?   (1 mark)  

Show Answers Only

 `a=4.5`

Show Worked Solution

`text(If)\ int_-a^a f(x)\ dx =0`

 ♦♦♦ A devilish 1-mark question mid-paper that had a mean mark of just 12%. 
MARKER’S COMMENT: The fact that this question was worth only 1 mark means that it is not necessary for students to show any detailed working.

`text(We know the area below the curve)`

`text(and above the)\ x text(-axis = area above the)`

`text(curve and below the)\ x text(-axis.)`

`text(By inspection, we can see)`

`int_-3^-1 f(x)\ dx=0\ \ text(and)\ \  int_2^3 f(x)\ dx= 0`

`text(We need)\ int_3^a f(x)\ dx + int_-a^-3 f(x)\ dx`  `=-3`
`text(because)\ \  int_-1^2 f(x)\ dx=3`   

 
`=>\ text(S)text(ince areas have height = 1, each)`

`text(must be 1.5 units wide.)`

`:. a = 4.5`