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QCAA Mathematical Methods — Introduction to integration Questions & Answers

Q3

2024

QCAA

Paper 2

1 mark

Q3

1 mark

The derivative of the function $f(x)$ is given by $f'(x) = \sin(2x)$ . It is known that $f\left(\frac{\pi}{2}\right) = 4$ .
Determine $f(x)$ .

A

$-\cos(2x) + 3$

B

$\cos(2x) + 5$

C

$-\frac{1}{2}\cos(2x) + 3.5$

D

$\frac{1}{2}\cos(2x) + 4.5$

Reveal Answer

A

$-\cos(2x) + 3$

This option fails to apply the reverse chain rule (u-substitution). The integral of $\sin(kx)$ is $-\frac{1}{k}\cos(kx)$ , so you must divide by the coefficient $2$ .

B

$\cos(2x) + 5$

This option uses the wrong sign for the antiderivative and misses the chain rule factor. The integral of $\sin(x)$ is $-\cos(x)$ , not $\cos(x)$ , and the result must be divided by $2$ .

C

$-\frac{1}{2}\cos(2x) + 3.5$

Correct Answer

Integrating $f'(x) = \sin(2x)$ yields $f(x) = -\frac{1}{2}\cos(2x) + C$ . Using the condition $f(\frac{\pi}{2}) = 4$ , we solve $4 = -\frac{1}{2}\cos(\pi) + C$ to find $C = 3.5$ .

D

$\frac{1}{2}\cos(2x) + 4.5$

This option has the wrong sign for the cosine term. Since the derivative of $\cos(x)$ is $-\sin(x)$ , the antiderivative of $\sin(x)$ must be negative.

Q1

2025

QCAA

Paper 2

1 mark

Q1

1 mark

Determine $\int 10.4x^3 \, dx$

A

$2.6x^4 + c$

B

$6.4x^4 + c$

C

$14.4x^4 + c$

D

$41.6x^4 + c$

Reveal Answer

A

$2.6x^4 + c$

Correct Answer

Correct. Using the power rule for integration, we increase the exponent by 1 and divide the coefficient by the new exponent: $\frac{10.4}{3+1}x^{3+1} + c = 2.6x^4 + c$ .

B

$6.4x^4 + c$

Incorrect. This option incorrectly subtracts the new exponent (4) from the coefficient 10.4 instead of dividing by it.

C

$14.4x^4 + c$

Incorrect. This option incorrectly adds the new exponent (4) to the coefficient 10.4 instead of dividing by it.

D

$41.6x^4 + c$

Incorrect. This option incorrectly multiplies the coefficient 10.4 by the new exponent (4) instead of dividing by it, confusing the integration rule with the differentiation rule.

Q2

2024

VCAA

Paper 2

1 mark

Q2

1 mark

A function $g : R \rightarrow R$ has the derivative $g'(x) = x^3 - x$ .

Given that $g(0) = 5$ , the value of $g(2)$ is

A

2

B

3

C

5

D

7

Reveal Answer

A

2

Incorrect. Evaluating the antiderivative $g(x) = \frac{x^4}{4} - \frac{x^2}{2} + 5$ at $x=2$ yields 7, not 2.

B

3

Incorrect. This might result from ignoring the constant of integration $C=5$ and calculating $g(2) = 4 - 2 = 2$ , then making an arithmetic error.

C

5

Incorrect. This is the value of the initial condition $g(0)$ , not the requested value $g(2)$ .

D

7

Correct Answer

Correct. Integrating $g'(x)$ gives $g(x) = \frac{x^4}{4} - \frac{x^2}{2} + C$ . Substituting $g(0) = 5$ gives $C = 5$ , and evaluating $g(2)$ yields $\frac{16}{4} - \frac{4}{2} + 5 = 7$ .

Q17

2024

QCAA

Paper 1

3 marks

Q17

3 marks

A community group that uses social media created a new post on the internet on a day when they had 1000 members. The rate of change in their number of members (members/day) is given by $f'(t) = 3e^{0.5t}$ , where $t$ represents days after the new post.
Determine the time it will take for the community group to achieve seven times the initial number of members. Express your answer in the form $a \ln(b)$ .

Reveal Answer

7 times the members is 7000.
Let m be the time when 7000 members is reached.
The required change in members is 6000.
$6000 = \int_0^m 3e^{0.5t} dt$
$= \left[ 3 \times \frac{1}{0.5}e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5t} \right]_0^m$
$= \left[ 6e^{0.5m} - 6e^0 \right]$
$= \left[ 6e^{0.5m} - 6 \right]$
$6000 = 6e^{0.5m} - 6$
$6006 = 6e^{0.5m}$
$\frac{6006}{6} = e^{0.5m}$
$0.5m = \ln 1001$
$m = 2\ln 1001 \text{ days}$

Marking Criteria

Descriptor	Marks
Correctly uses the initial conditions to determine the increase	1
Correctly determines the integral	1
Determines the number of days required	1

Q3

2020

VCAA

Paper 2

1 mark

Q3

1 mark

Let $f'(x) = \frac{2}{\sqrt{2x - 3}}$ .

If $f(6) = 4$ , then

A

$f(x) = 2\sqrt{2x - 3}$

B

$f(x) = \sqrt{2x - 3} - 2$

C

$f(x) = 2\sqrt{2x - 3} - 2$

D

$f(x) = \sqrt{2x - 3} + 2$

E

$f(x) = \sqrt{2x - 3}$

Reveal Answer

A

$f(x) = 2\sqrt{2x - 3}$

Incorrect. While this function has the correct derivative, it evaluates to $f(6) = 6$ , meaning the constant of integration $C$ was incorrectly assumed to be $0$ .

B

$f(x) = \sqrt{2x - 3} - 2$

Incorrect. The antiderivative of $f'(x)$ is $2\sqrt{2x - 3} + C$ . This option incorrectly integrates the function, missing a factor of $2$ .

C

$f(x) = 2\sqrt{2x - 3} - 2$

Correct Answer

Correct. Integrating $f'(x)$ using u-substitution ( $u = 2x - 3$ ) gives $f(x) = 2\sqrt{2x - 3} + C$ . Solving $f(6) = 4$ yields $C = -2$ .

D

$f(x) = \sqrt{2x - 3} + 2$

Incorrect. This function has the incorrect derivative $f'(x) = \frac{1}{\sqrt{2x-3}}$ and evaluates to $f(6) = 5$ , failing both conditions.

E

$f(x) = \sqrt{2x - 3}$

Incorrect. This function has the incorrect derivative $f'(x) = \frac{1}{\sqrt{2x-3}}$ and evaluates to $f(6) = 3$ , failing both conditions.

Q11

2025

QCAA

Paper 1

5 marks

Q11

Determine the following integrals.

Q11a

1 mark

$\int \frac{1}{x} dx$

Reveal Answer

\int \frac{1}{x} dx = \ln(x) + c

Marking Criteria

Descriptor	Marks
correctly applies the integration rule	1

Q11b

1 mark

$\int e^{4x} dx$

Reveal Answer

\int e^{4x} dx = \frac{1}{4}e^{4x} + c

Marking Criteria

Descriptor	Marks
correctly integrates the exponential term	1

Q11c

3 marks

$\int x(x^3 + 8) dx$

Reveal Answer

\begin{align*} \int x(x^3 + 8)dx &= \int (x^4 + 8x)dx\\ &= \frac{x^5}{5} + 4x^2 + c \end{align*}

Marking Criteria

Descriptor	Marks
correctly expands the brackets	1
integrates the power of 4 term	1
integrates the linear term	1

Q2

2021

VCAA

Paper 1

2 marks

Q2

2 marks

Let $f'(x) = x^3 + x$ .

Find $f(x)$ given that $f(1) = 2$ .

Reveal Answer

$f(x) = \int (x^3 + x) dx$
$f(x) = \frac{x^4}{4} + \frac{x^2}{2} + c$
$f(1) = \frac{1}{4} + \frac{1}{2} + c = 2$
$c = \frac{5}{4}$
$\therefore f(x) = \frac{x^4}{4} + \frac{x^2}{2} + \frac{5}{4}$

Marking Criteria

Descriptor	Marks
Finds the correct general antiderivative, including the constant of integration $c$ (e.g., $f(x) = \frac{x^4}{4} + \frac{x^2}{2} + c$ ).	1
Correctly substitutes the given condition $f(1) = 2$ to evaluate the constant of integration and states the final function $f(x) = \frac{x^4}{4} + \frac{x^2}{2} + \frac{5}{4}$ .	1

Q5

2024

QCAA

Paper 1

1 mark

Q5

1 mark

Determine $\int_a^b 2\cos(x)dx$ , where $a = \frac{\pi}{3}$ and $b = \frac{\pi}{2}$

A

$1 - \frac{\sqrt{3}}{2}$

B

$\frac{\sqrt{3}}{2} - 1$

C

$2 - \sqrt{3}$

D

$\sqrt{3} - 2$

Reveal Answer

A

$1 - \frac{\sqrt{3}}{2}$

This option neglects the constant factor of 2. It represents the value of $\int_{\pi/3}^{\pi/2} \cos(x)dx$ , which is $1 - \frac{\sqrt{3}}{2}$ .

B

$\frac{\sqrt{3}}{2} - 1$

This result comes from missing the factor of 2 and reversing the order of subtraction (calculating lower limit minus upper limit).

C

$2 - \sqrt{3}$

Correct Answer

The antiderivative of $2\cos(x)$ is $2\sin(x)$ . Applying the Fundamental Theorem of Calculus yields $2\sin(\frac{\pi}{2}) - 2\sin(\frac{\pi}{3}) = 2(1) - 2(\frac{\sqrt{3}}{2}) = 2 - \sqrt{3}$ .

D

$\sqrt{3} - 2$

This answer results from swapping the limits of integration or subtracting in the wrong order ( $F(a) - F(b)$ ), yielding the negative of the correct answer.

Q14

2025

VCAA

Paper 2

1 mark

Q14

1 mark

Let $f$ be the probability density function for a continuous random variable $X$ , where

f(x) = \begin{cases} k\sin(x) & 0 \le x < \frac{\pi}{4} \\ k\cos(x) & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ 0 & \text{otherwise} \end{cases}

and $k$ is a positive real number.

The value of $k$ is

A

$\frac{1}{\sqrt{2}}$

B

$\frac{1}{2 - \sqrt{2}}$

C

$\sqrt{2} + 2$

D

$2 - \sqrt{2}$

Reveal Answer

A

$\frac{1}{\sqrt{2}}$

Incorrect. This value does not make the total area under the probability density function equal to 1, likely resulting from an error in evaluating the trigonometric integrals.

B

$\frac{1}{2 - \sqrt{2}}$

Correct Answer

Correct. For $f(x)$ to be a valid probability density function, its integral over all $x$ must equal 1. Evaluating $\int_0^{\pi/4} k\sin(x)dx + \int_{\pi/4}^{\pi/2} k\cos(x)dx = 1$ yields $k(2 - \sqrt{2}) = 1$ , which gives $k = \frac{1}{2 - \sqrt{2}}$ .

C

$\sqrt{2} + 2$

Incorrect. This might result from incorrectly rationalizing the denominator or making an arithmetic error when solving $k(2 - \sqrt{2}) = 1$ .

D

$2 - \sqrt{2}$

Incorrect. This is the value of the integral when $k=1$ . Since the total area must be 1, $k$ must be the reciprocal of this value.

Q6

2023

QCAA

Paper 2

1 mark

Q6

1 mark

$\int_a^{5a} \frac{1}{x+a} dx$ , $a \neq 0$ is

A

1.7918

B

1.6094

C

1.3863

D

1.0986

Reveal Answer

A

1.7918

This value corresponds to $\ln(6)$ . This error typically occurs if you evaluate the antiderivative at the upper limit ( $x=5a$ ) but forget to subtract the evaluation at the lower limit ( $x=a$ ).

B

1.6094

This value corresponds to $\ln(5)$ . This is incorrect; the integration results in the natural log of the ratio of the bounds adjusted by $a$ , which is $\ln(3)$ , not $\ln(5)$ .

C

1.3863

This value corresponds to $\ln(4)$ . The correct evaluation of the definite integral yields $\ln(3)$ .

D

1.0986

Correct Answer

The antiderivative is $\ln|x+a|$ . Applying the limits yields $\ln(5a+a) - \ln(a+a) = \ln(6a) - \ln(2a) = \ln(\frac{6a}{2a}) = \ln(3) \approx 1.0986$ .

Q18

2025

QCAA

Paper 1

6 marks

Q18

6 marks

Two objects are launched simultaneously from different positions and travel along the same straight-line path. The objects are launched towards each other with the same initial speed.

The first object's displacement (m) from the origin is given by $d_1 = \frac{1}{3}t^3 - \frac{1}{2}t^2 + kt$ , where $t$ is the time (s) since the objects were launched and $k$ is a constant, $k \neq 0$ . The second object is moving with a constant acceleration of $4 \text{ m s}^{-2}$ .

The second object changes its direction, and at time $t = 1 \text{ s}$ the objects have equal velocities and continue to travel in the same direction.

Compared to the first object, how much further does the second object travel between $t = 1 \text{ s}$ and the next time the objects have equal velocities?

Reveal Answer

Finding velocity of the first object:

v_1(t) = \frac{d}{dt}(\frac{1}{3}t^3 - \frac{1}{2}t^2 + kt) = t^2 - t + k

Find the rule for the velocity for the second object:

v_2(t) = \int 4 dt = 4t + c

Given condition for

\begin{align*} v_1(1) &= v_2(1)\\ 1^2 - 1 + k &= 4 + c\\ c &= k - 4 \end{align*}

Find the time when the velocities are the same.

\begin{align*} v_1(t) &= v_2(t)\\ t^2 - t + k &= 4t + k - 4\\ t^2 - 5t + 4 &= 0\\ (t - 1)(t - 4) &= 0\\ \therefore t &= 1, 4 \end{align*}

Need the time different to the given 1 s, so $t = 4$ .

As objects travel in the one direction between given times, the distance travelled is equal to the displacement. Find the difference between the objects' travelling distances.

\begin{align*} \text{displacement} &= \int \text{velocity} \, dt\\ d_2 - d_1 &= \int_1^4 (v_2 - v_1) dt\\ &= \int_1^4 ((4t - 2) - (t^2 - t + 2)) dt\\ &= \int_1^4 (-t^2 + 5t - 4) dt\\ &= \left[ \frac{-1}{3}t^3 + \frac{5}{2}t^2 - 4t \right]_1^4\\ &= \frac{-1}{3} \times (4)^3 + \frac{5}{2} \times (4)^2 - 4 \times 4 - (\frac{-1}{3} + \frac{5}{2} - 4)\\ &= \frac{-63}{3} + 28 - \frac{5}{2}\\ &= -21 + 28 - 2.5\\ &= 4.5 \end{align*}

Marking Criteria

Descriptor	Marks
correctly determines the equation for the first object's velocity	1
correctly determines the equation for the second object's velocity including the constant c	1
determine the relationship between constants $k$ and $c$	1
determines the second time when the two velocities are the same	1
uses a suitable method for determining the difference between distances the objects have travelled in the given time interval	1
determines the difference between distances of the two objects	1

Q5

2023

VCAA

Paper 1

4 marks

Q5a

1 mark

Evaluate $\int_0^{\frac{\pi}{3}} \sin(x) dx$ .

Reveal Answer

$\frac{1}{2}$

Marking Criteria

Descriptor	Marks
Evaluates the definite integral to find the correct answer of $\frac{1}{2}$	1

Q5b

3 marks

Hence, or otherwise, find all values of $k$ such that $\int_0^{\frac{\pi}{3}} \sin(x) dx = \int_k^{\frac{\pi}{2}} \cos(x) dx$ , where $-3\pi < k < 2\pi$ .

Reveal Answer

\int_k^{\frac{\pi}{2}} \cos(x)dx = [\sin(x)]_k^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(k) = 1 - \sin(k)

Using part a.,

$\frac{1}{2} = 1 - \sin(k)$

$\sin(k) = \frac{1}{2}$

$k = \frac{-11\pi}{6}, \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$

Marking Criteria

Descriptor	Marks
Correctly evaluates the right-hand side integral to obtain $1 - \sin(k)$	1
Equates the evaluated integral to the answer from part a. and simplifies to $\sin(k) = \frac{1}{2}$	1
Finds all correct values of $k$ within the given domain: $\frac{-11\pi}{6}, \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$	1

Q17

2022

QCAA

Paper 2

4 marks

Q17

4 marks

A snail is travelling along a straight path from point $A$ . The snail's velocity (cm min $^{-1}$ ) is modelled by $v(t)=1.4\ln\left(1+t^2\right)$ , where $t$ is time (in minutes) for $0 \leq t \leq 15$ .

An ant passes point $A$ 12 minutes after the snail and follows the snail's path. The ant moves with a constant acceleration of 2 cm min $^{-2}$ and passes the snail at $t=15$ minutes.

Determine the ant's velocity at point $A$ .

Reveal Answer

Total displacement of the snail
$\int_0^{15} 1.4 \ln(1 + t^2) dt = 76.0431$ cm

Velocity of the ant $= \int 2 dt$
$= 2t + c$

Displacement $_{ant \ from \ 12 \ to \ 15min} =$ Displacement $_{snail \ from \ 0 \ to \ 15min}$
$\therefore \int_{12}^{15} 2t + c = 76.0431$

Solving numerically on GDC
$c = -1.6523$

Therefore, velocity of ant at $t = 12$
$= 2 \times 12 - 1.6523$
$= 22.3477$ cm min $^{-1}$ along the ant's path.

Marking Criteria

Descriptor	Marks
correctly determines the total displacement of the snail	1
establishes an equation linking the ant and the snail	1
determines constant	1
determines velocity of ant	1

Q11

2021

VCAA

Paper 2

1 mark

Q11

1 mark

If $\int_0^a f(x)dx = k$ , then $\int_0^a (3f(x) + 2)dx$ is

A

$3k + 2a$

B

$3k$

C

$k + 2a$

D

$k + 2$

E

$3k + 2$

Reveal Answer

A

$3k + 2a$

Correct Answer

Correct. By the linearity of integrals, $\int_0^a (3f(x) + 2)dx = 3\int_0^a f(x)dx + \int_0^a 2dx = 3k + 2a$ .

B

$3k$

Incorrect. This evaluates $\int_0^a 3f(x)dx$ but completely ignores the integral of the constant $2$ .

C

$k + 2a$

Incorrect. This correctly evaluates the integral of $2$ as $2a$ , but forgets to multiply the integral of $f(x)$ by the constant factor $3$ .

D

$k + 2$

Incorrect. This forgets to multiply the integral of $f(x)$ by $3$ and incorrectly evaluates $\int_0^a 2dx$ as $2$ instead of $2a$ .

E

$3k + 2$

Incorrect. This correctly multiplies the integral of $f(x)$ by $3$ , but incorrectly evaluates $\int_0^a 2dx$ as $2$ instead of $2a$ .

Q1

2025

QCAA

Paper 1

4 marks

Q1

4 marks

Determine the value of $\int_{1}^{2} 3x^2 \, dx$

A

9

B

7

C

6

D

5

Reveal Answer

A

9

Incorrect. This result likely comes from adding the evaluated bounds ( $2^3 + 1^3 = 9$ ) instead of subtracting them, which violates the Fundamental Theorem of Calculus.

B

7

Correct Answer

Correct. The antiderivative of $3x^2$ is $x^3$ . Evaluating this from 1 to 2 using the Fundamental Theorem of Calculus gives $2^3 - 1^3 = 8 - 1 = 7$ .

C

6

Incorrect. This value might come from evaluating the derivative $6x$ at $x=1$ , rather than finding the definite integral.

D

5

Incorrect. This is a miscalculation. Remember to find the antiderivative $x^3$ and evaluate $F(2) - F(1)$ .

QCAA Mathematical Methods Introduction to integration

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