How many QCAA Specialist Mathematics questions cover Applications of integral calculus?

AusGrader has 66 QCAA Specialist Mathematics questions on Applications of integral calculus, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Applications of integral calculus Questions & Answers

Q6

2025

VCAA

Paper 1

4 marks

Q6

4 marks

Find the volume of the solid of revolution formed when the area between the curve

y = \sqrt{\frac{\arctan(x)}{1+x^2}}

and the $x$ -axis from $x = 1$ to $x = \sqrt{3}$ is rotated about the $x$ -axis.

Give your answer in the form $\frac{a\pi^b}{c}$ , where $a, b, c \in Z^+$ .

Reveal Answer

$V = \pi \int_1^{\sqrt{3}} \frac{\arctan(x)}{1+x^2} dx$

Method 1 (substitution)
$u = \arctan(x), \ du = \frac{dx}{1+x^2}$ .

\begin{align*} V &= \pi \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} u \ du\\ &= \pi \left[ \frac{u^2}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\\ &= \frac{\pi^3}{2} \left( \frac{1}{9} - \frac{1}{16} \right)\\ &= \frac{7\pi^3}{288} \end{align*}

Marking Criteria

Descriptor	Marks
Sets up the correct definite integral for the volume of revolution, including the factor of $\pi$	1
Identifies the correct substitution $u = \arctan(x)$ and corresponding $du = \frac{dx}{1+x^2}$	1
Correctly changes the limits of integration to $\frac{\pi}{4}$ and $\frac{\pi}{3}$ and finds the antiderivative	1
Evaluates the integral to obtain the correct final answer $\frac{7\pi^3}{288}$	1

Q13

2020

QCAA

Paper 1

4 marks

Q13

4 marks

The expected value of an exponential random variable $X$ with parameter $\lambda > 0$ can be determined using the rule

$E(X) = \int_0^\infty x\lambda e^{-\lambda x} dx$

Use integration by parts to determine $E(X)$ .
Express your answer in simplest form.

Reveal Answer

$E(X) = \int_0^\infty x \lambda e^{-\lambda x} dx$
$\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$

$u = x \quad \frac{du}{dx} = 1$
$\frac{dv}{dx} = \lambda e^{-\lambda x} \quad v = -e^{-\lambda x}$

$E(X) = -xe^{-\lambda x}|_0^\infty + \int_0^\infty e^{-\lambda x} dx$
$= 0 + \int_0^\infty e^{-\lambda x} dx$
$= \frac{e^{-\lambda x}}{-\lambda}|_0^\infty$
$= 0 - \frac{1}{-\lambda}$
$= \frac{1}{\lambda}$

Marking Criteria

Descriptor	Marks
correctly determines $\frac{du}{dx}$ and $v$	1
substitutes into the integration by parts rule	1
calculates $-xe^{-\lambda x}\|_0^\infty$ to equal 0	1
shows that $E(X) = \frac{1}{\lambda}$	1

Q5

2024

VCAA

Paper 1

3 marks

Q5

3 marks

The curve with equation $y = \sqrt{k - \frac{1}{x^2}}$ , for $1 \le x \le \frac{k}{2}$ where $k > 2$ , is rotated about the $x$ -axis to form a solid of revolution that has volume $\frac{7\pi}{2}$ units $^3$ .

Show that $k$ satisfies the equation $k^3 - 2k^2 - 9k + 4 = 0$ .

Reveal Answer

The volume $V$ is given by $V = \pi \int_1^k \left(k - \frac{1}{x^2}\right) dx$ .

$V = \pi \int_1^k \left(k - \frac{1}{x^2}\right) dx= \pi \left[kx + \frac{1}{x}\right]_1^k= \pi \left(\frac{1}{2}k^2 + \frac{2}{k} - k - 1\right)$

$\pi \left(\frac{1}{2}k^2 + \frac{2}{k} - k - 1\right) = \frac{7\pi}{2}\implies k^3 - 2k^2 - 9k + 4 = 0$

Marking Criteria

Descriptor	Marks
Sets up the correct definite integral for the volume	1
Correctly integrates the expression and substitutes the limits	1
Equates the evaluated integral to $\frac{7\pi}{2}$ and correctly rearranges to show $k^3 - 2k^2 - 9k + 4 = 0$	1

Q9

2024

QCAA

Paper 2

1 mark

Q9

1 mark

Random variable $X$ has an exponential distribution with the probability density function

$f(x) = \begin{cases} \frac{1}{5}e^{-\frac{x}{5}}, & x \ge 0 \\ 0, & \text{otherwise} \end{cases}$

Given that $P(0 \le X \le k) = 0.5$ , determine $k$ .

A

0.10

B

0.69

C

2.03

D

3.47

Reveal Answer

A

0.10

This value is incorrect because substituting $k=0.10$ into the cumulative distribution function yields a probability of approximately $0.02$ , not $0.5$ .

B

0.69

This option represents $\ln(2)$ , which would be the median for a standard exponential distribution with $\lambda=1$ , failing to account for the parameter $\lambda = 1/5$ .

C

2.03

This value is incorrect; calculating the cumulative probability $1 - e^{-2.03/5}$ results in approximately $0.33$ , which is less than the required $0.5$ .

D

3.47

Correct Answer

Using the CDF $F(k) = 1 - e^{-k/5}$ , we set $1 - e^{-k/5} = 0.5$ . Solving for $k$ gives $k = -5\ln(0.5) \approx 3.47$ .

Q17

2021

QCAA

Paper 1

7 marks

Q17

7 marks

The area between the graphs of the functions $y = 4x$ and $y = 2x^2$ is rotated about the $y$ -axis to form a solid of revolution with a volume of $V$ units $^3$ .

Determine the exact value of $V$ .

Reveal Answer

Finding the points of intersection of the two functions $y = 4x$ and $y = 2x^2$
$4x = 2x^2$
$2x^2 - 4x = 0 \Rightarrow 2x(x - 2) = 0 \Rightarrow x = 0$ and $x = 2$
When $x = 0, y = 0$
When $x = 2, y = 8$

Rearranging the two functions in the form $x = f(y)$ and $x = g(y)$
$x = \frac{y}{4}$ and $x = \pm \sqrt{\frac{y}{2}}$
Finding volume of revolution between curves
$V = \left| \pi \int_a^b [f(y)]^2 - [g(y)]^2 dy \right|$
$= \left| \pi \int_0^8 \frac{y}{2} - \frac{y^2}{16} dy \right|$
$= \pi \left| \frac{y^2}{4} - \frac{y^3}{48} \right|_0^8$
$= \pi \left| (16 - \frac{32}{3}) - (0) \right|$
$= \frac{16\pi}{3}$

Marking Criteria

Descriptor	Marks
correctly uses simultaneous equations to establish an equation in one unknown	1
correctly determines y-coordinates of the points of intersection	1
correctly determines functions in the form $x = f(y)$	1
determines expression to represent the volume between the two curves	1
integrates expression	1
determines (positive) value of $V$ in terms of $\pi$	1
shows logical organisation, communicating key steps to at least the start of finding the volume of revolution	1

Q17

2022

QCAA

Paper 1

5 marks

Q17

5 marks

The region between the $x$ -axis and the curve of the function $y = 1 + \sin(2x)$ for $0 \le x \le \frac{\pi}{2}$ is rotated about the $x$ -axis to form a solid of revolution.

Determine the volume of this solid. Express your answer in simplest form.

Reveal Answer

$V = \pi \int_{a}^{b} y^2 dx$
$= \pi \int_{0}^{\frac{\pi}{2}} (1 + \sin(2x))^2 dx$
$= \pi \int_{0}^{\frac{\pi}{2}} 1 + 2\sin(2x) + \sin^2(2x) dx$
$= \pi \int_{0}^{\frac{\pi}{2}} 1 + 2\sin(2x) + \frac{1}{2}(1 - \cos(4x)) dx$
$= \pi \int_{0}^{\frac{\pi}{2}} (\frac{3}{2} + 2\sin(2x) - \frac{1}{2}\cos(4x)) dx$
$= \pi [\frac{3}{2}x - \cos(2x) - \frac{1}{8}\sin(4x)]_{0}^{\frac{\pi}{2}}$
$= \pi [(\frac{3\pi}{4} - \cos(\pi) - \frac{1}{8}\sin(2\pi)) - (0 - \cos(0) - 0)]$
$= \pi (\frac{3\pi}{4} + 2)$ units $^3$

Marking Criteria

Descriptor	Marks
correctly substitutes into the appropriate volume of a solid of revolution rule	1
expands an integrand	1
uses a suitable double-angle identity to enable an integration process to be completed	1
integrates an expression	1
determines volume in simplest form	1

Q8

2020

VCAA

Paper 1

5 marks

Q8

5 marks

Find the volume, $V$ , of the solid of revolution formed when the graph of $y = 2\sqrt{\frac{x^2 + x + 1}{(x+1)(x^2+1)}}$ is rotated about the $x$ -axis over the interval $\left[0, \sqrt{3}\right]$ . Give your answer in the form $V = 2\pi(\log_e(a) + b)$ , where $a, b \in R$ .

Reveal Answer

Many students identified the correct form of the partial fraction decomposition for the integrand:
$\frac{A}{x+1} + \frac{Bx+C}{x^2+1}$

This led to the integral $2\pi\int_0^{\sqrt{3}} \left(\frac{1}{x+1} + \frac{x}{x^2+1} + \frac{1}{x^2+1}\right) dx$

The answer is:
$2\pi\left(\log_e(2\sqrt{3} + 2) + \frac{\pi}{3}\right)$

Marking Criteria

Descriptor	Marks
Sets up the correct definite integral for the volume of the solid of revolution, $V = \pi \int_0^{\sqrt{3}} 4\frac{x^2 + x + 1}{(x+1)(x^2+1)} dx$	1
Correctly decomposes the integrand using partial fractions (e.g., identifying the form $\frac{A}{x+1} + \frac{Bx+C}{x^2+1}$ and finding the constants)	1
Finds the correct antiderivative, including both logarithmic and inverse tangent terms (e.g., $\log_e(x+1) + \frac{1}{2}\log_e(x^2+1) + \arctan(x)$ )	1
Correctly substitutes the limits of integration $0$ and $\sqrt{3}$ into the antiderivative	1
Obtains the correct final answer in the required form, $2\pi\left(\log_e(2\sqrt{3} + 2) + \frac{\pi}{3}\right)$	1

Q9

2023

QCAA

Paper 2

1 mark

Q9

1 mark

The time in minutes between the arrival of customers at a certain shop is assumed to be a random variable $X$ with an exponential distribution that has the probability density function

$f(x) = \begin{cases} 0.12e^{-0.12x}, & x \geq 0 \\ 0, & \text{otherwise} \end{cases}$

A customer arrives at the shop. The probability that the next customer arrives within 30 to 60 seconds, to the nearest percent, is

A

3%

B

5%

C

7%

D

11%

Reveal Answer

A

3%

This value is too low. The probability is calculated by integrating the probability density function over the interval $[0.5, 1]$ minutes.

B

5%

Correct Answer

Since $X$ is in minutes, convert the time interval 30 to 60 seconds to $0.5$ to $1$ minute. The probability is $\int_{0.5}^{1} 0.12e^{-0.12x} dx = e^{-0.06} - e^{-0.12} \approx 0.0548$ , which is approximately $5\%$ .

C

7%

This value is incorrect. Ensure you converted seconds to minutes correctly ( $30\text{ sec} = 0.5\text{ min}$ , $60\text{ sec} = 1\text{ min}$ ) before applying the exponential distribution formula.

D

11%

This value is incorrect. It may result from failing to convert the time units from seconds to minutes or using the wrong limits of integration.

Q4

2022

QCAA

Paper 2

1 mark

Q4

1 mark

The time taken for students to answer questions in a class is assumed to be a random variable $X$ with an exponential distribution that has the probability density function

$f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \ge 0 \\ 0, & \text{otherwise} \end{cases}$

The mean of $X$ is $\frac{1}{\lambda}$ .
The mean length of time taken for students to answer questions in this class is 15 seconds.
The probability that the next question in this class is answered between 8 seconds and 17 seconds is

A

0.05

B

0.12

C

0.22

D

0.26

Reveal Answer

A

0.05

This value is incorrect. The probability is calculated by integrating the PDF from 8 to 17, which yields a significantly higher value than 0.05.

B

0.12

This value is incorrect. It does not match the result derived from the exponential distribution formula $P(a < X < b) = e^{-\lambda a} - e^{-\lambda b}$ .

C

0.22

This value is incorrect. While closer to the answer, the calculation using $\lambda = \frac{1}{15}$ results in approximately 0.26.

D

0.26

Correct Answer

Given the mean is 15, the rate parameter is $\lambda = \frac{1}{15}$ . The probability is $P(8 < X < 17) = \int_{8}^{17} \frac{1}{15}e^{-\frac{x}{15}} dx = e^{-\frac{8}{15}} - e^{-\frac{17}{15}} \approx 0.587 - 0.322 = 0.265$ .

Q14

2023

QCAA

Paper 2

4 marks

Q14

4 marks

At a certain location, a biologist measures the width of a river to be 12 m. She also records the depth of the river at regular 2 m interval widths as shown.

Width (m)	0	2	4	6	8	10	12
Depth (m)	0.52	2.15	3.70	4.27	3.32	1.28	0.59

The biologist estimates the cross-sectional area of the river at this location to be $15 \text{ m}^2$ .

Use Simpson's rule to evaluate the reasonableness of this estimation. Justify your area calculation and decision regarding reasonableness using mathematical reasoning.

Reveal Answer

Use Simpson's rule to estimate the cross-sectional area of the river.

Using the given data: $w = 2$ .

$\text{Area} \approx \frac{w}{3}(y_0 + 4(y_1 + y_3 + \dots) + 2(y_2 + y_4 + \dots) + y_n)$
$\approx \frac{2}{3}(0.52 + 4(2.15 + 4.27 + 1.28) + \dots$
$2(3.70 + 3.32) + 0.59)$

$\text{Area} \approx 30.63\text{m}^2$

The estimate of the area is less than half of the value obtained using Simpson's rule, so it is not reasonable.

Marking Criteria

Descriptor	Marks
correctly identifies the interval width	1
justifies the area calculation by substituting the depth values of the data into Simpson's rule	1
calculates area	1
states and justifies a decision regarding the reasonableness of the estimation using mathematical reasoning	1

Q10

2025

VCAA

Paper 2

1 mark

Q10

1 mark

The region bounded by the curve given by $y = 3 \cos^{-1}(x)$ , for $0 \leq y \leq a$ , where $a > 0$ , and the line $x = 0$ is rotated about the $y$ -axis to form a solid of revolution. The volume of the solid is $\frac{\pi(4\pi + 3\sqrt{3})}{8}$ .

The value of $a$ is

A

$\frac{\pi}{4}$

B

$\frac{\pi}{3}$

C

$\frac{\pi}{2}$

D

$\pi$

Reveal Answer

A

$\frac{\pi}{4}$

Substituting $a = \frac{\pi}{4}$ into the evaluated volume integral $V = \frac{\pi}{2} \left( a + \frac{3}{2} \sin\left(\frac{2a}{3}\right) \right)$ yields $\frac{\pi}{2}\left(\frac{\pi}{4} + \frac{3}{4}\right)$ , which does not match the given volume.

B

$\frac{\pi}{3}$

Substituting $a = \frac{\pi}{3}$ into the evaluated volume integral yields $\frac{\pi}{2}\left(\frac{\pi}{3} + \frac{3}{2}\sin\left(\frac{2\pi}{9}\right)\right)$ , which does not match the given volume.

C

$\frac{\pi}{2}$

Substituting $a = \frac{\pi}{2}$ into the evaluated volume integral yields $\frac{\pi}{2}\left(\frac{\pi}{2} + \frac{3\sqrt{3}}{4}\right)$ , which is close but results in $\frac{\pi(2\pi + 3\sqrt{3})}{8}$ instead of the required volume.

D

$\pi$

Correct Answer

Using the disk method, the volume is $V = \pi \int_0^a \cos^2\left(\frac{y}{3}\right) dy = \frac{\pi}{2} \left(a + \frac{3}{2} \sin\left(\frac{2a}{3}\right)\right)$ . Setting $a = \pi$ gives exactly $\frac{\pi(4\pi + 3\sqrt{3})}{8}$ .

Q14

2020

QCAA

Paper 2

5 marks

Q14

The time, $t$ , (months) that it takes before a phone owner cracks the screen on their phone can be modelled by an exponentially distributed random variable

$f(t) = \begin{cases} 0.16e^{-0.16t}, & t \ge 0 \\ 0, & \text{otherwise} \end{cases}$

Q14a

1 mark

Show that $f(t)$ is a probability density function.

Reveal Answer

Using integration facility of GDC
$\int_0^\infty 0.16e^{-0.16t} dt = 1$

Marking Criteria

Descriptor	Marks
correctly substitutes the given information into a definite integral to show that f(t) is a probability density function	1

Q14b

2 marks

Determine the probability that a phone owner cracks the screen on their phone within 1 year.

Reveal Answer

1 year = 12 months
$P(0 < x < 12) = \int_0^{12} 0.16e^{-0.16t} dt$
Using integration facility of GDC
$\approx 0.85$

Marking Criteria

Descriptor	Marks
correctly represents the required probability as a definite integral	1
determines the probability	1

Q14c

2 marks

Three-quarters of phone owners take between 1 and $m$ months before they crack the screen on their phone.

Determine the value of $m$ .

Reveal Answer

$P(1 < x < m) = 0.75$
$\int_1^m 0.16e^{-0.16t} dt = 0.75$
Using solve facility of GDC or otherwise
$m \approx 14.26$ months

Marking Criteria

Descriptor	Marks
correctly establishes definite integral equation	1
solves equation to determine m	1

Q7

2020

VCAA

Paper 1

5 marks

Q7

Consider the function defined by

f(x) = \begin{cases} mx + n, & x < 1 \\ \frac{4}{1+x^2}, & x \ge 1 \end{cases}

where $m$ and $n$ are real numbers.

Q7a

2 marks

Given that $f(x)$ and $f'(x)$ are continuous over $R$ , show that $m = -2$ and $n = 4$ .

Reveal Answer

$m = -2, n = 4$

Marking Criteria

Descriptor	Marks
Establishes the equation $m + n = 2$ using the continuity of $f(x)$ at $x = 1$	1
Calculates the derivative of $\frac{4}{1+x^2}$ and evaluates it at $x = 1$ to show $m = -2$ , and concludes $n = 4$	1

Q7b

3 marks

Find the area enclosed by the graph of the function, the $x$ -axis and the lines $x = 0$ and $x = \sqrt{3}$ .

Reveal Answer

$\int_0^1 -2x + 4 dx + \int_1^{\sqrt{3}} \frac{4}{1+x^2} dx = \left[-x^2 + 4x\right]_0^1 + \left[4\arctan(x)\right]_1^{\sqrt{3}}$
$= (3 - 0) + 4\left(\frac{\pi}{3} - \frac{\pi}{4}\right)$
$= 3 + \frac{\pi}{3}$

Marking Criteria

Descriptor	Marks
Sets up the correct expression for the area, such as $\int_0^1 (-2x + 4) dx + \int_1^{\sqrt{3}} \frac{4}{1+x^2} dx$ (or uses the area of a trapezium for the first part)	1
Finds the correct antiderivatives, $\left[-x^2 + 4x\right]$ and $\left[4\arctan(x)\right]$	1
Evaluates the definite integrals to find the correct final area of $3 + \frac{\pi}{3}$	1

QCAA Specialist Mathematics Applications of integral calculus

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