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

AusGrader has 48 QCAA Specialist Mathematics questions on Applications of integral calculus, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Applications of integral calculus Questions & Answers | AusGrader

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

Correct Answer

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

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

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 $V$ .

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.

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

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.

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

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

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

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

QCAA Specialist Mathematics Applications of integral calculus

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