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Calculus, 2ADV 2026 HSC 8 MC

A section of the graph of  \(y=f(x)\)  for  \(x \ge 0\)  is shown, where \(f(x)\) is an odd function.
 

Four integrals based on this graph are given.

\(P=\displaystyle\int_{0}^{a} f(x)\,dx \qquad R=\displaystyle\int_{-a}^{1} f(x)\,dx\)

\(Q=\displaystyle\int_{-a}^{a} f(x)\,dx \qquad S=\displaystyle\int_{-a}^{0} f(x)\,dx\)

Which of the following options identifies the integrals in ascending numerical order?

  1. \(P, R, S, Q\)
  2. \(P, S, R, Q\)
  3. \(S, Q, R, P\)
  4. \(S, R, Q, P\)
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\(C\)

Show Worked Solution

\(f(x)\ \text{is odd}\ \Rightarrow\ Q=\displaystyle\int_{-a}^{a} f(x)\,dx=0\)

\(P>0\ \ \text{(area above}\ x\text{-axis} > \text{area below)}\)

\(S=\displaystyle\int_{-a}^{0} f(x)\,dx=-P<0\)

\(R=\displaystyle\int_{-a}^{a} f(x)\,dx-\displaystyle\int_{1}^{a} f(x)\,dx=0-\displaystyle\int_{1}^{a} f(x)\,dx>0\)

\(R\ \text{equals the small area between}\ x=1\ \text{and}\ x=a\ \Rightarrow\ R<P\)

\(\therefore\ S<Q<R<P\)

\(\Rightarrow C\)

Calculus, 2ADV C4 2025 MET2 17 MC

Given that  \(\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x\), the graph of  \(y=f(x)\) could be
 

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\(A\)

Show Worked Solution

\(\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x\)

\(\text{To be true,} \ \displaystyle \int_2^3 f(x)\, d x<0 \ \ \text{(“net” area below} \ x \text{-axis)}\)

\(\Rightarrow A\)

♦ Mean mark 38%.

Calculus, 2ADV C4 EQ-Bank 26

  1. The graph of \(f(x)\) is drawn below
     

  1. Evaluate \(\displaystyle \int_0^6 f(x)\, d x\)   (2 marks)

  2. Evaluate \(\displaystyle \int_0^6[f(x)-3]\, d x\)   (2 marks)

  3. Evaluate \(\displaystyle \int_4^6 f^{\prime}(x)\, d x\)   (1 mark)

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a.    \(8 \dfrac{1}{2}\)

b.    \(-9 \dfrac{1}{2}\)

c.    \(-3\)

Show Worked Solution

a.
         

\(\displaystyle \int_0^6 f(x)\) \(=\text{net area above the}\ x \text{-axis }\)
  \(=\text{Area 1}+ \text{Area 3}-\text{Area 2}\)
  \(=\left(6+1 \dfrac{1}{2}\right)+3-2\)
  \(=8 \dfrac{1}{2}\)

 

b.     \(f(x)-3 \ \ \text{shifts graph (above) 3 units lower:}\)
 

\(\displaystyle\int_0^6[f(x)-3] \ \ \text{will be negative (areas below \(x\)-axis)}\)
  \(=-\left(1 \dfrac{1}{2}+5+3\right)\)
  \(=-9 \dfrac{1}{2}\)

 

c.     \(\displaystyle \int_4^6 f^{\prime}(x) d x\) \(=[f(x)]_4^6\)
    \(=f(6)-f(4)\)
    \(=0-3\)
    \(=-3\)

Calculus, 2ADV C4 EQ-Bank 30

Evaluate  \(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x\).   (3 marks)

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\(\pi\)

Show Worked Solution

\(\sqrt{4-x^2} \ \ \text{is a semicircle, centre}\ \ (0,0) \ \ \text {and radius 2.}\)
 

STRATEGY: This integral is beyond 2ADV integration techniques. An alternate strategy is required.
\(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x \ \) \( \ =\ \text{Shaded area (above)}\)
  \(=\dfrac{1}{4} \times \pi+2^2\)
  \(=\pi\)

Calculus, 2ADV C4 2024 MET2 4*

If \( { \displaystyle \int_a^b f(x) d x=-5 } \)  and \( { \displaystyle \int_a^c f(x) d x=3 } \), where  \(a<b<c\).

Find  \( { \displaystyle \int_b^c 2 f(x) d x } \).   (2 marks)

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\(16\)

Show Worked Solution
\({ \displaystyle \int_a^b f(x) d x} +{ \displaystyle \int_b^c f(x) d x}\) \(={ \displaystyle \int_a^c f(x) d x} \)
\(-5+{ \displaystyle \int_b^c f(x) d x}\) \(=3\)
\({ \displaystyle \int_b^c f(x) d x}\) \(=3+5=8\)

 
\(\therefore{ \displaystyle \int_b^c 2f(x) d x}=16\)

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`
Show Answers Only

`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`
Show Answers Only

`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)

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a.    `t > 5\ \ text(seconds)`

b.    `7.4\ \ text(seconds)`

c.    
       

Show Worked Solution

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

`:. t > 5\ \ text(seconds)`
  

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

`text(Consider the displacement from)\ B\ text(to)\ D:`

`text(Since the area below the graph from)\ B\ text(to)\ C\ text(equals )`

`text(the area above the graph from)\ C\ text(to)\ D,\ text(there is no )`

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

 

`text(Consider)\ 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.)`
     

c.   

     

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`  
Show Answers Only

`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)  

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`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 and)`

`text(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`