How many QCAA Specialist Mathematics questions cover Integration techniques?

AusGrader has 88 QCAA Specialist Mathematics questions on Integration techniques, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Integration techniques 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

Q2

2021

VCAA

Paper 2

1 mark

Q2

1 mark

The implied domain of the function with rule $f(x) = \cos^{-1}(\log_e(bx))$ , $b > 0$ is

A

$(0, 1]$

B

$[1, e]$

C

$\left[\frac{1}{b}, \frac{e}{b}\right]$

D

$\left[\frac{1}{b}, \frac{e^\pi}{b}\right]$

E

$\left[\frac{1}{be}, \frac{e}{b}\right]$

Reveal Answer

A

$(0, 1]$

This option is incorrect because it does not properly account for the domain of the inverse cosine function, which requires the argument to be between $-1$ and $1$ .

B

$[1, e]$

This would be the domain if $b=1$ and the inequality was incorrectly set to $0 \le \log_e(x) \le 1$ instead of $-1 \le \log_e(x) \le 1$ .

C

$\left[\frac{1}{b}, \frac{e}{b}\right]$

This incorrectly assumes the domain of $\cos^{-1}(u)$ is $[0, 1]$ instead of $[-1, 1]$ , which would lead to solving $0 \le \log_e(bx) \le 1$ .

D

$\left[\frac{1}{b}, \frac{e^\pi}{b}\right]$

This incorrectly uses the range of $\cos^{-1}(u)$ , which is $[0, \pi]$ , instead of its domain $[-1, 1]$ to set up the inequality $0 \le \log_e(bx) \le \pi$ .

E

$\left[\frac{1}{be}, \frac{e}{b}\right]$

Correct Answer

The domain of $\cos^{-1}(u)$ is $[-1, 1]$ , so we must have $-1 \le \log_e(bx) \le 1$ . Exponentiating both sides gives $e^{-1} \le bx \le e$ , and dividing by $b > 0$ yields $x \in \left[\frac{1}{be}, \frac{e}{b}\right]$ .

Q2

2021

VCAA

Paper 1

3 marks

Q2

3 marks

Evaluate $\int_0^1 \frac{2x+1}{x^2+1}\,dx$ .

Reveal Answer

Write the integrand as the sum of two rational functions:
$\int_0^1 \frac{2x}{x^2+1} dx + \int_0^1 \frac{1}{x^2+1} dx = \left[\log_e(x^2+1)\right]_0^1 + \left[\arctan(x)\right]_0^1$

Answer is:
$\log_e(2) + \frac{\pi}{4}$

Marking Criteria

Descriptor	Marks
Splits the integrand into $\frac{2x}{x^2+1}$ and $\frac{1}{x^2+1}$ or correctly finds one antiderivative	1
Correctly finds both antiderivatives, obtaining $\log_e(x^2+1) + \arctan(x)$	1
Correctly evaluates the definite integral to obtain the final answer $\log_e(2) + \frac{\pi}{4}$	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

Q2

2020

VCAA

Paper 2

1 mark

Q2

1 mark

A function $f$ has the rule $f(x)=\left|b\cos^{-1}(x)-a\right|$ , where $a>0$ , $b>0$ and $a<\frac{b\pi}{2}$ .

The range of $f$ is

A

$[-a,\,b\pi-a]$

B

$[0,\,b\pi-a]$

C

$[a,\,b\pi-a]$

D

$[0,\,b\pi+a]$

E

$[a-b\pi,\,a]$

Reveal Answer

A

$[-a,\,b\pi-a]$

The absolute value function ensures all outputs are non-negative, so the range cannot include negative values like $-a$ .

B

$[0,\,b\pi-a]$

Correct Answer

The range of $\cos^{-1}(x)$ is $[0, \pi]$ , so the range of $b\cos^{-1}(x)-a$ is $[-a, b\pi-a]$ . Since $a < \frac{b\pi}{2}$ , we know $b\pi-a > a$ . Taking the absolute value makes the minimum $0$ (since the interval crosses zero) and the maximum $b\pi-a$ .

C

$[a,\,b\pi-a]$

This assumes the minimum value is $|-a|=a$ . However, since the inner function's range $[-a, b\pi-a]$ includes $0$ , the minimum of the absolute value is $0$ .

D

$[0,\,b\pi+a]$

The maximum value of the inner function is $b(\pi)-a = b\pi-a$ . The value $b\pi+a$ would only be possible if the inner function was $b\cos^{-1}(x)+a$ .

E

$[a-b\pi,\,a]$

The absolute value function cannot produce negative values, and these bounds do not correctly reflect the transformation of the arccosine range.

Q2

2024

QCAA

Paper 1

1 mark

Q2

1 mark

Given that $\frac{A}{x-2} + \frac{3}{x} = \frac{x-6}{x(x-2)}$ , determine the value of $A$ .

A

$-4$

B

$-2$

C

$2$

D

$4$

Reveal Answer

A

$-4$

Substituting $A=-4$ and combining the fractions yields a numerator of $-4x + 3(x-2) = -x-6$ , which does not match the required numerator $x-6$ .

B

$-2$

Correct Answer

Multiplying the equation by the common denominator $x(x-2)$ gives $Ax + 3(x-2) = x-6$ . Expanding and comparing coefficients of $x$ leads to $A+3=1$ , so $A=-2$ .

C

$2$

Substituting $A=2$ results in a combined numerator of $2x + 3(x-2) = 5x-6$ , which is incorrect.

D

$4$

Substituting $A=4$ results in a combined numerator of $4x + 3(x-2) = 7x-6$ , which is incorrect.

Q9

2022

VCAA

Paper 1

4 marks

Q9

4 marks

Given that $f'(x) = \frac{\cos(2x)}{\sin^3(2x)}$ and $f\left(\frac{\pi}{8}\right) = \frac{3}{4}$ , find $f(x)$ .

Reveal Answer

With $u = \sin(2x)$ ,

$\int \frac{\cos(2x)}{\sin^3(2x)} dx = \frac{1}{2}\int \frac{du}{u^3}$
$= -\frac{1}{4}u^{-2} + c$
$= -\frac{1}{4} \cdot \frac{1}{\sin^2(2x)} + c$

When $x = \frac{\pi}{8}$ :

$-\frac{1}{4} \cdot 2 + c = \frac{3}{4}$
$\Rightarrow c = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$

Therefore

$f(x) = \frac{-1}{4\sin^2(2x)} + \frac{5}{4} = -\frac{1}{4}\text{cosec}^2(2x) + \frac{5}{4}$

Marking Criteria

Descriptor	Marks
Sets up the integral using an appropriate substitution, such as $u = \sin(2x)$	1
Correctly integrates to find the general antiderivative, $-\frac{1}{4\sin^2(2x)} + c$	1
Substitutes the given condition $f\left(\frac{\pi}{8}\right) = \frac{3}{4}$ to solve for the constant of integration	1
States the correct final function $f(x) = -\frac{1}{4\sin^2(2x)} + \frac{5}{4}$ or equivalent	1

Q6

2025

QCAA

Paper 1

1 mark

Q6

1 mark

At time $t$ , a particle travels with a velocity of $v = \left(\frac{2}{1+t^2}\right)\hat{i} - 2t\hat{j}$ .

Determine a general expression for the position vector, $r$ , of the particle during this motion.

A

$r = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c$

B

$r = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c$

C

$r = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c$

D

$r = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c$

Reveal Answer

A

$r = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c$

This is incorrect because the integral of the $\hat{j}$ component, $-2t$ , is $-t^2$ , not $-2$ .

B

$r = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c$

Correct Answer

This is correct. The position vector is the integral of the velocity vector with respect to time. Integrating $\frac{2}{1+t^2}$ yields $2\tan^{-1}(t)$ , and integrating $-2t$ yields $-t^2$ .

C

$r = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c$

This is incorrect because both components are integrated improperly. The integral of $\frac{2}{1+t^2}$ is $2\tan^{-1}(t)$ , and the integral of $-2t$ is $-t^2$ .

D

$r = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c$

This is incorrect because the integral of the $\hat{i}$ component, $\frac{2}{1+t^2}$ , is $2\tan^{-1}(t)$ , not $\frac{1}{2}\tan^{-1}(t)$ .

Q1

2020

SCSA

Paper 1

3 marks

Q1

3 marks

Evaluate exactly $\int_0^\pi (4\cos^2 x - \sin x) \, dx$ .

Reveal Answer

\begin{align*} \int_0^\pi (4\cos^2 x - \sin x) dx &= \int_0^\pi (2(1+\cos 2x) - \sin x) dx\\ &= \int_0^\pi (2 + 2\cos 2x - \sin x) dx \quad ... (1)\\ &= \left[ 2x + \sin 2x + \cos x \right]_0^\pi\\ &= (2\pi + 0 - 1) - (0 + 0 + 1)\\ &= 2\pi - 2 \end{align*}

Marking Criteria

Descriptor	Marks
uses the cosine double angle identity to obtain integrand (1)	1
anti-differentiates the integrand correctly	1
evaluates the definite integral correctly	1

Q15

2021

QCAA

Paper 1

4 marks

Q15

4 marks

Use partial fractions to determine $\int \frac{4x-17}{x^2 - x - 6} dx$ , where $x \in R, x \neq -2, x \neq 3$ .

Express your answer in the form $\ln|f(x)| + c$ .

Reveal Answer

$\frac{4x - 17}{x^2 - x - 6} = \frac{A}{(x + 2)} + \frac{B}{(x - 3)}$
$= \frac{A(x - 3) + B(x + 2)}{(x + 2)(x - 3)}$
$x = 3: -5 = 5B \Rightarrow B = -1$
$x = -2: -25 = -5A \Rightarrow A = 5$
$\int \frac{4x - 17}{x^2 - x - 6} dx = \int (\frac{5}{(x + 2)} + \frac{-1}{(x - 3)}) dx$
$= 5\ln|x + 2| - \ln|x - 3|$
$= \ln|x + 2|^5 - \ln|x - 3|$
$= \ln \left| \frac{(x + 2)^5}{x - 3} \right| + c$

Marking Criteria

Descriptor	Marks
correctly factorises the denominator to establish the form of the partial fraction decomposition	1
determines values of A and B	1
determines indefinite integral of the fraction	1
determines expression in the form $\ln\|f(x)\|$	1

Q4

2022

SCSA

Paper 1

8 marks

Q4a

3 marks

Function $f(x) = \frac{5(x + 1)}{(x - 1)(x^2 + 3x + 1)}$ can be expressed in the form $\frac{a}{x - 1} + \frac{bx + c}{x^2 + 3x + 1}$ .

Determine the value of the constants $a$ , $b$ and $c$ .

Reveal Answer

\begin{aligned} \frac{5x+5}{(x-1)(x^2+3x+1)} &= \frac{a(x^2+3x+1) + (x-1)(bx+c)}{(x-1)(x^2+3x+1)} \\ &= \frac{(a+b)x^2 + (3a-b+c)x + (a-c)}{(x-1)(x^2+3x+1)} \end{aligned}

Equating coefficients:

\begin{align*} a+b &= 0\\ 3a-b+c &= 5\\ a-c &= 5 \end{align*}

Solving gives $a=2$ , $b=-2$ , $c=-3$

\begin{align*} \text{i.e. }\frac{5(x+1)}{(x-1)(x^2+3x+1)} = \frac{2}{x-1} - \frac{(2x+3)}{x^2+3x+1} \end{align*}

Marking Criteria

Descriptor	Marks
forms the correct expression for the equivalent numerator	1
equates coefficients correctly to form 3 linear equations	1
solves correctly to determine $a$ , $b$ and $c$	1

Q4b

5 marks

Hence determine $\int \frac{10x + 10}{(x - 1)(x^2 + 3x + 1)} \, dx$ .

Reveal Answer

\begin{aligned} \int \frac{10x+10}{(x-1)(x^2+3x+1)} dx &= 2 \int \frac{5x+5}{(x-1)(x^2+3x+1)} dx \\ &= \int \frac{4}{x-1} - \frac{2(2x+3)}{x^2+3x+1} dx \\ &= 4\ln|x-1| - 2\ln|x^2+3x+1| + k \\ &= \ln\left( \frac{(x-1)^4}{(x^2+3x+1)^2} \right) + k \end{aligned}

Marking Criteria

Descriptor	Marks
expresses the given integrand as double $f(x)$	1
writes the integrand correctly in terms of the partial fractions	1
anti-differentiates $\frac{a}{x-1}$ correctly using the absolute value of a natural logarithm	1
anti-differentiates $\frac{bx+c}{x^2+3x+1}$ correctly	1
uses a constant of integration	1

Q3

2022

SCSA

Paper 1

5 marks

Q3

5 marks

By using one or more of the following identities:

$\cos^2 x + \sin^2 x = 1$
$\cos 2x = \cos^2 x - \sin^2 x$
$\sin 2x = 2 \sin x \cos x$

evaluate exactly $\int_0^{\frac{\pi}{2}} (\sin x + \cos x)^2 \, dx$ .

Reveal Answer

\begin{aligned} \int_0^{\frac{\pi}{2}} (\sin x + \cos x)^2 dx &= \int_0^{\frac{\pi}{2}} (\sin^2 x + 2\sin x \cos x + \cos^2 x) dx \\ &= \int_0^{\frac{\pi}{2}} ((\sin^2 x + \cos^2 x) + 2\sin x \cos x) dx \\ &= \int_0^{\frac{\pi}{2}} (1 + \sin 2x) dx \\ &= \left[ x - \frac{\cos 2x}{2} \right]_0^{\frac{\pi}{2}} \\ &= \left( \frac{\pi}{2} - \frac{\cos \pi}{2} \right) - \left( 0 - \frac{\cos 0}{2} \right) \\ &= \frac{\pi}{2} + \frac{1}{2} - \left( -\frac{1}{2} \right) \\ &= \frac{\pi}{2} + 1 \end{aligned}

Marking Criteria

Descriptor	Marks
expands the integrand correctly	1
uses the Pythagorean identity $\sin^2 x + \cos^2 x = 1$	1
uses double angle identity for $\sin 2x$	1
anti-differentiates the trigonometric function correctly	1
evaluates correctly using exact trigonometric values	1

Q7

2022

VCAA

Paper 2

1 mark

Q7

1 mark

Using the substitution $u = 1 + e^x$ , $\int_0^{\log_e 2} \frac{1}{1 + e^x} dx$ can be expressed as

A

$\int_0^{\log_e 2} \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

B

$\int_2^3 \left( \frac{1}{u} - \frac{1}{u - 1} \right) du$

C

$\int_1^3 \left( \frac{1}{u} - \frac{1}{u - 1} \right) du$

D

$\int_2^3 \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

E

$\int_2^{1 + e^2} \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

Reveal Answer

A

$\int_0^{\log_e 2} \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

This option is incorrect because the limits of integration were not updated. When substituting $u = 1 + e^x$ , the new limits should be $u(0) = 2$ and $u(\log_e 2) = 3$ .

B

$\int_2^3 \left( \frac{1}{u} - \frac{1}{u - 1} \right) du$

This option has the correct limits but the wrong partial fraction decomposition. The integrand $\frac{1}{u(u-1)}$ decomposes to $\frac{1}{u-1} - \frac{1}{u}$ , not the reverse.

C

$\int_1^3 \left( \frac{1}{u} - \frac{1}{u - 1} \right) du$

This option is incorrect because the lower limit was evaluated incorrectly ( $1 + e^0 = 2$ , not $1$ ) and the partial fraction decomposition has the wrong signs.

D

$\int_2^3 \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

Correct Answer

Correct. Substituting $u = 1 + e^x$ gives $dx = \frac{du}{u-1}$ , changing the integrand to $\frac{1}{u(u-1)} = \frac{1}{u-1} - \frac{1}{u}$ . The new limits are $u(0) = 1 + e^0 = 2$ and $u(\log_e 2) = 1 + 2 = 3$ .

E

$\int_2^{1 + e^2} \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$

This option is incorrect because the upper limit was evaluated improperly. When $x = \log_e 2$ , $u = 1 + e^{\log_e 2} = 1 + 2 = 3$ , not $1 + e^2$ .

Q3

2021

QCAA

Paper 1

1 mark

Q3

1 mark

An object has a velocity $v(t) = e^{-2t}\hat{i} + \left(\frac{1}{t}\right)\hat{k}$ , where $t$ represents time ( $t > 0$ ).

The displacement $r(t)$ of the object could be

A

$-2e^{-2t}\hat{i} + \ln(t)\hat{k}$

B

$-2e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

C

$-\frac{1}{2}e^{-2t}\hat{i} + \ln(t)\hat{k}$

D

$-\frac{1}{2}e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

Reveal Answer

A

$-2e^{-2t}\hat{i} + \ln(t)\hat{k}$

This option incorrectly differentiates the $\hat{i}$ component instead of integrating it. The integral of $e^{-2t}$ is $-\frac{1}{2}e^{-2t}$ , not $-2e^{-2t}$ .

B

$-2e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

This option represents the acceleration (derivative of velocity) rather than displacement. Both components were differentiated: $\frac{d}{dt}(e^{-2t}) = -2e^{-2t}$ and $\frac{d}{dt}(t^{-1}) = -t^{-2}$ .

C

$-\frac{1}{2}e^{-2t}\hat{i} + \ln(t)\hat{k}$

Correct Answer

Displacement is found by integrating the velocity function with respect to time: $\int e^{-2t} dt = -\frac{1}{2}e^{-2t}$ and $\int \frac{1}{t} dt = \ln(t)$ .

D

$-\frac{1}{2}e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

The $\hat{k}$ component is incorrect because it represents the derivative of $\frac{1}{t}$ (which is $-\frac{1}{t^2}$ ) rather than the integral (which is $\ln(t)$ ).

Q3

2023

SCSA

Paper 1

5 marks

Q3

5 marks

Using the substitution $x = 119u + 1$ , evaluate exactly $\int_{1}^{120} \left( 2 + 4 \left( \frac{x + 118}{119} \right)^3 \right) dx$ .

Reveal Answer

Using $x = 119u+1 \quad \therefore dx = 119\,du$
$x+118 = 119u+119$
$\therefore \frac{x+118}{119} = \frac{119u+119}{119} = u+1$

$x$	1	120
$u$	0	1

\begin{align*} \int_1^{120} \left( 2+4\left(\frac{x+118}{119}\right)^3 \right) dx &= \int_0^1 \left( 2+4(u+1)^3 \right) . 119\,du\\ &= 119 \left[ 2u+(u+1)^4 \right]_0^1\\ &= 119 \left[ (2+2^4)-(0+1^4) \right]\\ &= 119 [18-1]\\ &= 119 \times 17\\ &= 2023 \end{align*}

Marking Criteria

Descriptor	Marks
writes $dx$ correctly in terms of $du$	1
changes the limits correctly	1
simplifies the integrand correctly in terms of the variable $u$	1
anti-differentiates correctly	1
evaluates the definite integral correctly	1

QCAA Specialist Mathematics Integration techniques

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