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AusGrader has 102 SCSA Mathematics Specialist questions on Integration and applications of integration, all with instant AI grading and detailed marking feedback.

WACE Mathematics Specialist — Integration and applications of integration Questions & Answers

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

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 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

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.

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

Q18

2024

QCAA

Paper 1

6 marks

Q18

6 marks

A random variable $X$ has a probability density function given by

$f(x) = \begin{cases} k\sin^{-1}(x), & 0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$

where $k$ is a positive constant.

Determine the value of $k$ .

Reveal Answer

Using a property of a pdf

$\int_0^1 k \sin ^{-1}(x) dx=1 \quad \dots (1)$

Consider $\int k \sin ^{-1}(x) dx$

Using integration by parts

$\int u \frac{dv}{dx} dx=u v-\int v \frac{du}{dx} dx$

Let $u=\sin ^{-1}(x) \quad \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}$

Let $\frac{dv}{dx}=k \quad v=k x$

$\int k \sin ^{-1}(x) dx=k x \sin ^{-1}(x)-\int \frac{k x}{\sqrt{1-x^2}} dx$

Using substitution for the developed integral $\int \frac{k x}{\sqrt{1-x^2}} dx$

Let $u=1-x^2$

$\frac{du}{dx}=-2x$

$\int \frac{k x}{\sqrt{1-x^2}} dx=-\frac{k}{2} \int \frac{-2 x}{\sqrt{1-x^2}} dx$

$=-\frac{k}{2} \int \frac{1}{\sqrt{u}} du$

$\int \frac{k x}{\sqrt{1-x^2}} dx=-\frac{k}{2} \int u^{-\frac{1}{2}} du$

$=-k u^{\frac{1}{2}}$

$=-k \sqrt{1-x^2}+c$

$\therefore \int k \sin ^{-1} x dx=k x \sin ^{-1}(x)+k \sqrt{1-x^2}+c$

Substituting into (1)

$\int_0^1 k \sin ^{-1}(x) dt=1$

$\left[k\left(x \sin ^{-1}(x)+\sqrt{1-x^2}\right)\right]_0^1=1$

$k\left(\left(\sin ^{-1}(1)+\sqrt{1-1}\right)-\left(0+\sqrt{1-0}\right)\right)=1$

$k\left(\frac{\pi}{2}-1\right)=1$

$k=\frac{2}{\pi-2}$

Marking Criteria

Descriptor	Marks
Correctly uses a suitable pdf property	1
Correctly determines both results required to progress integration by parts	1
Uses integration by parts	1
Uses a suitable substitution method to progress the developed integral within the use of integration by parts	1
Determines a general result for the required integral	1
Determines value of k	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

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

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

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

2020

QCAA

Paper 1

7 marks

Q17

7 marks

Determine the smallest positive value of $a$ given

$\int_{-a}^{a} 1 + \left(\frac{\sec(2x) + \tan(2x)}{\text{cosec}(2x) + 1}\right)^2 dx = 1$

Reveal Answer

$\int_{-a}^a 1 + \left(\frac{\sec(2x) + \tan(2x)}{\text{cosec}(2x) + 1}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\frac{1}{\cos(2x)} + \frac{\sin(2x)}{\cos(2x)}}{\frac{1}{\sin(2x)} + 1}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\frac{1 + \sin(2x)}{\cos(2x)}}{\frac{1 + \sin(2x)}{\sin(2x)}}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\sin(2x)}{\cos(2x)}\right)^2 dx = 1$
$\int_{-a}^a 1 + (\tan(2x))^2 dx = 1$
$\int_{-a}^a \sec^2(2x)dx = 1$
$\frac{1}{2}\tan(2x)\Big|_{-a}^a = 1$
$\frac{1}{2}(\tan(2a) - \tan(-2a)) = 1$
$\frac{1}{2}(\tan(2a) + \tan(2a)) = 1$
$\frac{1}{2}(2\tan(2a)) = 1 \Rightarrow 2a = \tan^{-1}(1)$
$2a = \frac{\pi}{4} \Rightarrow a = \frac{\pi}{8}$

Marking Criteria

Descriptor	Marks
correctly expresses the expression inside the brackets of the integrand in terms of $\sin(2x)$ and $\cos(2x)$	1
correctly simplifies the expression inside the brackets of the integrand	1
uses a suitable Pythagorean identity to express the integrand as a single trigonometric expression	1
determines the definite integral equation	1
substitutes limits of integration	1
solves equation to determine the smallest positive value of $a$	1
shows logical organisation communicating key steps	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 $1 + e^2$ .

Q4

2024

SCSA

Paper 1

6 marks

Q4a

2 marks

Given that $\frac{a}{x + 1} + \frac{b}{(x + 1)^2} = \frac{5x + 3}{(x + 1)^2}$ , determine the values for $a$ and $b$ .

Reveal Answer

$\frac{a}{x+1} + \frac{b}{(x+1)^2} = \frac{a(x+1)+b}{(x+1)^2} = \frac{ax+(a+b)}{(x+1)^2}$
Equating coefficients:
$a+b = 3$
$a = 5$
$b = -2$
Solving gives $a = 5, \ b = -2$

Marking Criteria

Descriptor	Marks
forms the equivalence of numerators correctly	1
solves for $a, b$ correctly	1

Q4b

4 marks

Hence determine $\int \frac{5x + 3}{(x + 1)^2} \, dx$ .

Reveal Answer

\begin{align*} \int \frac{5x+3}{(x+1)^2} \, dx &= \int \frac{5}{x+1} - \frac{2}{(x+1)^2} \, dx\\ &= 5\ln|x+1| + \frac{2}{x+1} + k \end{align*}

Marking Criteria

Descriptor	Marks
re-writes the integrand correctly in terms of the partial fractions	1
anti-differentiates the $(x+1)^{-1}$ term correctly using the absolute value of a natural logarithm	1
anti-differentiates the $2(x+1)^{-2}$ term correctly	1
uses a constant of integration	1

Q13

2022

QCAA

Paper 1

6 marks

Q13a

4 marks

Use partial fractions to determine $\int \frac{22}{(2x-3)(x+4)} dx$

Reveal Answer

$\frac{22}{(2x-3)(x+4)} = \frac{A}{2x-3} + \frac{B}{x+4}$
$\therefore 22 = A(x+4) + B(2x-3)$
Let $x = \frac{3}{2}: A(\frac{3}{2}+4) = 22 \Rightarrow A = 4$
Let $x = -4: B(2 \times -4 - 3) = 22 \Rightarrow B = -2$
$\int \frac{22}{2x^2+5x-12} dx = \int \frac{4}{2x-3} dx + \int \frac{-2}{x+4} dx$
$= 2 \int \frac{2}{2x-3} dx - 2 \int \frac{1}{x+4} dx$
$= 2 \ln|2x-3| - 2 \ln|x+4| + c$

Marking Criteria

Descriptor	Marks
correctly sets up the partial fractions	1
determines value of A	1
determines value of B	1
determines an expression for the indefinite integral	1

Q13b

2 marks

Use the result from Question 13a) to determine $\int_{-3}^{0} \frac{22}{(2x-3)(x+4)} dx$
Express your answer in simplest form.

Reveal Answer

$\int_{-3}^{0} \frac{22}{(2x-3)(x+4)} dx$
$= [2 \ln|2x-3| - 2 \ln|x+4|]_{-3}^{0}$
$= ((2 \ln|-3| - 2 \ln|4|) - (2 \ln|-9| - 2 \ln|1|))$
$= 2(\ln(3) - \ln(4) - \ln(9)) = 2 \ln(\frac{3}{9 \times 4})$
$= 2 \ln(\frac{1}{12})$

Marking Criteria

Descriptor	Marks
substitutes limits of integration into the result from 13a)	1
expresses a definite integral value in simplest form	1

SCSA Mathematics Specialist Integration and applications of integration

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