How many SCSA Mathematics Methods questions cover Integrals?

AusGrader has 182 SCSA Mathematics Methods questions on Integrals, all with instant AI grading and detailed marking feedback.

WACE Mathematics Methods — Integrals Questions & Answers

Q13

2024

QCAA

Paper 1

6 marks

Q13a

3 marks

$F(x) = \int \left(4e^{2x} + \sin(2x)\right) dx$ . Use integration to determine $F(x)$ , if $F(0) = 5$ .

Reveal Answer

$F(x) = \int \left(4e^{2x} + \sin(2x)\right) dx$
$= \frac{4e^{2x}}{2} + -\frac{1}{2}\cos(2x) + c$
$= 2e^{2x} - \frac{1}{2}\cos(2x) + c$
$F(0) = 5$
$5 = 2 - \frac{1}{2} + c$
$c = 3\frac{1}{2}$
$F(x) = 2e^{2x} - \frac{1}{2}\cos(2x) + 3\frac{1}{2}$

Marking Criteria

Descriptor	Marks
Correctly integrates the exponential term	1
Correctly integrates the trigonometric term	1
Determines the constant of integration	1

Q13b

3 marks

If $\frac{dy}{dx} = \left(\frac{3x^7 - 2x}{x^4}\right)^2$ , determine $y$ .

Reveal Answer

$\frac{dy}{dx} = \left(\frac{3x^7 - 2x}{x^4}\right)^2$
$= \left(\frac{3x^7}{x^4} - \frac{2x}{x^4}\right)^2$
$= \left(3x^3 - \frac{2}{x^3}\right)^2$
$= (3x^3)^2 - 2(3x^3)\left(\frac{2}{x^3}\right) + \left(\frac{2}{x^3}\right)^2$
$= 9x^6 - 12 + \frac{4}{x^6}$
$= 9x^6 - 12 + 4x^{-6}$
$y = \int (9x^6 - 12 + 4x^{-6}) dx$
$y = \frac{9}{7}x^7 - 12x - \frac{4}{5}x^{-5} + c$

Marking Criteria

Descriptor	Marks
Correctly expands the squared bracket	1
Simplifies the expanded squared bracket into separate terms	1
Determines y in terms of x by integrating all terms	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

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

Q12

2021

QCAA

Paper 2

4 marks

Q12

The velocity function of an object in m s $^{-1}$ is given by $v(t) = \cos\left(6t + \frac{\pi}{2}\right) + 2, 0 \leq t \leq 5$ .
Initially, the object is at the origin.

Q12a

2 marks

Determine the displacement function.

Reveal Answer

$s(t) = \frac{1}{6}\sin\left(6t + \frac{\pi}{2}\right) + 2t + c$
Substituting $(0, 0)$
$0 = \frac{1}{6}\sin\left(\frac{\pi}{2}\right) + c$
$c = \frac{-1}{6}$
$s(t) = \frac{1}{6}\sin\left(6t + \frac{\pi}{2}\right) + 2t + \frac{-1}{6}$

Marking Criteria

Descriptor	Marks
Correctly determines the indefinite integral $s(t)$	1
Correctly determines the displacement function	1

Q12b

2 marks

What is the displacement of the object from the origin, in metres (m), after three seconds?

Reveal Answer

distance $= s(3) - s(0)$
$= \frac{1}{6}\sin\left(18 + \frac{\pi}{2}\right) + 6 - \frac{1}{6} - (\frac{1}{6}\sin\left(\frac{\pi}{2}\right) - \frac{1}{6})$
$= 5.943$ m

Marking Criteria

Descriptor	Marks
Establishes an expression for the distance travelled	1
Determines distance travelled	1

Q11

2022

VCAA

Paper 2

1 mark

Q11

1 mark

If $\frac{d}{dx}(x \cdot \sin(x)) = \sin(x) + x \cdot \cos(x)$ , then $\frac{1}{k} \int x \cos(x) dx$ is equal to

A

$k \left( x \cdot \sin(x) - \int \sin(x) dx \right) + c$

B

$\frac{1}{k} x \cdot \sin(x) - \int \sin(x) dx + c$

C

$\frac{1}{k} \left( x \cdot \sin(x) - \int \sin(x) dx \right) + c$

D

$\frac{1}{k} (x \cdot \sin(x) - \sin(x)) + c$

E

$\frac{1}{k} \left( \int x \cdot \sin(x) dx - \int \sin(x) dx \right) + c$

Reveal Answer

A

$k \left( x \cdot \sin(x) - \int \sin(x) dx \right) + c$

This option incorrectly multiplies the result by $k$ instead of the required factor of $\frac{1}{k}$ .

B

$\frac{1}{k} x \cdot \sin(x) - \int \sin(x) dx + c$

This option fails to distribute the $\frac{1}{k}$ factor to the $\int \sin(x) dx$ term. The entire integral expression must be multiplied by $\frac{1}{k}$ .

C

$\frac{1}{k} \left( x \cdot \sin(x) - \int \sin(x) dx \right) + c$

Correct Answer

Integrating both sides of the given derivative yields $\int x \cos(x) dx = x \cdot \sin(x) - \int \sin(x) dx$ . Multiplying the entire expression by $\frac{1}{k}$ and adding the constant of integration $c$ gives this correct result.

D

$\frac{1}{k} (x \cdot \sin(x) - \sin(x)) + c$

This option incorrectly replaces $\int \sin(x) dx$ with $\sin(x)$ . The integral of $\sin(x)$ is actually $-\cos(x)$ .

E

$\frac{1}{k} \left( \int x \cdot \sin(x) dx - \int \sin(x) dx \right) + c$

This option incorrectly places an integral sign in front of $x \cdot \sin(x)$ . By the Fundamental Theorem of Calculus, integrating the derivative of $x \cdot \sin(x)$ simply yields $x \cdot \sin(x)$ .

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.

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 $\frac{16}{4} - \frac{4}{2} + 5 = 7$ .

Q11

2024

QCAA

Paper 2

4 marks

Q11

4 marks

State the trapezoidal rule and use it with six strips to determine an approximate value of the definite integral for the curve of $f(x) = 4(x-3)^2$ from $x = 0$ to $x = 3$ . Show all substitutions made into the rule.

Reveal Answer

$w=\dfrac{b-a}{n}$

$=\dfrac{3-0}{6}$

$=\dfrac{1}{2}$

$A\approx \dfrac{1}{2}w\left[f(x_0)+2f(x_1)+2f(x_2)+\ldots+2f(x_n)\right]$

$=\dfrac{1}{2}\times \dfrac{1}{2}\left[36+2\times 25+2\times 16+2\times 9+2\times 4+2\times 1+0\right]$

$=36.5\ \text{units}^2$

Marking Criteria

Descriptor	Marks
Correctly determines the rectangle width	1
Correctly states the trapezoidal rule	1
Substitutes appropriate values into the trapezoidal rule	1
Determines the approximate value of the definite integral	1

Q3

2022

QCAA

Paper 1

1 mark

Q3

1 mark

The area between the curve $y = 9 - x^2$ and the $x$ -axis is

A

12 units $^2$

B

18 units $^2$

C

36 units $^2$

D

54 units $^2$

Reveal Answer

A

12 units $^2$

This value is incorrect. It does not result from evaluating the definite integral of the function between the x-intercepts.

B

18 units $^2$

This is only half the area. It represents the integral from $x=0$ to $x=3$ , but the region extends symmetrically from $x=-3$ to $x=3$ .

C

36 units $^2$

Correct Answer

The curve intersects the x-axis at $x = \pm 3$ . The area is the integral $\int_{-3}^{3} (9 - x^2) dx = [9x - \frac{x^3}{3}]_{-3}^{3} = 18 - (-18) = 36$ .

D

54 units $^2$

This is the area of the bounding rectangle ( $width=6, height=9$ ). By Archimedes' property, the area under the parabola is $\frac{2}{3}$ of this rectangle ( $54 \times \frac{2}{3} = 36$ ).

Q8

2024

QCAA

Paper 2

1 mark

Q8

1 mark

Calculate the total enclosed area between the graph of $y=x^2-x-6$ and the $x$ -axis from $x=1$ to $x=5$

A

5.33

B

7.33

C

12.67

D

20.00

Reveal Answer

A

5.33

Incorrect. This is the value of the definite integral $\int_1^5 (x^2-x-6) dx$ , which treats the area below the axis as negative. To find the total geometric area, you must split the integral at the x-intercept ( $x=3$ ) and sum the absolute values.

B

7.33

Incorrect. This represents only the area of the region below the $x$ -axis, from $x=1$ to the root at $x=3$ . You must also add the area from $x=3$ to $x=5$ .

C

12.67

Incorrect. This represents only the area of the region above the $x$ -axis, from the root at $x=3$ to $x=5$ . You must also add the area from $x=1$ to $x=3$ .

D

20.00

Correct Answer

Correct. The function has a root at $x=3$ , so the area is calculated by splitting the integral: $|\int_1^3 (x^2-x-6) dx| + \int_3^5 (x^2-x-6) dx$ , which equals $|-7.33| + 12.67 = 20$ .

Q8

2022

VCAA

Paper 2

1 mark

Q8

1 mark

If $\int_0^b f(x) dx = 10$ and $\int_0^a f(x) dx = -4$ , where $0 < a < b$ , then $\int_a^b f(x) dx$ is equal to

A

$-6$

B

$-4$

C

$0$

D

$10$

E

$14$

Reveal Answer

A

$-6$

This incorrectly subtracts $10$ from $-4$ or adds the two values incorrectly. The correct operation is to subtract the integral from $0$ to $a$ from the integral from $0$ to $b$ .

B

$-4$

This is simply the given value for $\int_0^a f(x) dx$ , not the integral over the interval from $a$ to $b$ .

C

$0$

There is no mathematical justification for the integral evaluating to $0$ based on the given values.

D

$10$

This is the given value for $\int_0^b f(x) dx$ , which represents the area over the entire interval from $0$ to $b$ , not just from $a$ to $b$ .

E

$14$

Correct Answer

Using the additive property of definite integrals, $\int_a^b f(x) dx = \int_0^b f(x) dx - \int_0^a f(x) dx$ . Substituting the given values yields $10 - (-4) = 14$ .

Q19

2023

QCAA

Paper 2

6 marks

Q19

6 marks

Over a suitable domain, a hill has a cross-sectional area given by $\int h(x) dx = \frac{a}{b}e^{bx} + c$ , where:

$a, b$ and $c$ are constants, $b \neq 0$
$h(x)$ represents vertical distance (m), $x$ represents horizontal distance (m).

It is known that $h(0) = 1.22$ and $h(40) = 25$ .

Where the gradient of the hill is 0.86 there is a tree stump. A second tree stump is located further up the hill. The difference in hill gradient between the two tree stumps is 0.44.

A surveyor predicts that the vertical distance separating the two tree stumps is between 7.5 m and 8.5 m. Evaluate the reasonableness of this prediction.

Reveal Answer

The hill:
$\int h(x) dx = \frac{a}{b}e^{bx} + c$
Differentiating wrt x:
$h(x) = ae^{bx}$

Using $(0, 1.22)$
$1.22 = a \times e^0$
$a = 1.22$
$\therefore h(x) = 1.22e^{bx}$

Using $(40, 25)$
$25 = 1.22e^{40b}$
$\frac{25}{1.22} = e^{40b}$
$40b = \ln\left(\frac{25}{1.22}\right)$
$b = \frac{\ln\left(\frac{25}{1.22}\right)}{40}$
$b = 0.0755$
$\therefore h(x) = 1.22e^{0.0755x}$

The gradient of the hill:
$h'(x) = 0.09211e^{0.0755x}$

Determine the location of first tree stump:
Using $h'(x) = 0.86$
$0.86 = 0.09211e^{0.0755x}$
Solving for x:
$x = 29.5887$
$h(29.5887) = 11.3907$

Determine the location of the second tree stump:
The gradient is $0.86 + 0.44 = 1.3$
Using $h'(x) = 1.3$
$1.3 = 0.09211e^{0.0755x}$
Solving for x:
$x = 35.0614$
$h(35.0614) = 17.2185$

Vertical distance between the tree stumps:
$= h(35.0614) - h(29.5887)$
$= 17.2185 - 11.3907$
$= 5.8278$ m

Evaluation of the prediction: The vertical distance of 5.8278 m is NOT between 7.5 m and 8.5 m, so the prediction is NOT reasonable.

Marking Criteria

Descriptor	Marks
correctly determines the model for the hill with constants a and b found	1
differentiates h(x) to determine the gradient formula	1
determines the y-coordinate location of the first tree stump where the hill gradient is 0.86	1
determines the y-coordinate of the second tree stump	1
determines the vertical distance between the tree stumps	1
provides appropriate statement of reasonableness	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

2022

QCAA

Paper 1

6 marks

Q14

The rate that water fills an empty vessel is given by $\frac{dV}{dt} = 0.25e^{0.25t}$ (in litres per hour), $0 \le t \le 8\ln(6)$ , where $t$ is time (in hours).

Q14a

2 marks

Determine the function that represents the volume of water in the vessel (in litres).

Reveal Answer

$V = \int 0.25e^{0.25t} dt$
$= e^{0.25t} + c$
when $t = 0, V = 0$
$\therefore 0 = e^{0.25 \times 0} + c$
$\therefore c = -1$
$\therefore V = e^{0.25t} - 1$

Marking Criteria

Descriptor	Marks
correctly determines the integral of the function V(t)	1
determines the value of c	1

Q14b

2 marks

The vessel is full when $t = 8\ln(6)$ . Determine the volume of water, to the nearest litre, the vessel can hold when full.

Reveal Answer

$V(8\ln(6)) = e^{0.25 \times 8 \ln(6)} - 1$
$= 36 - 1$

$= 35$ litres

Marking Criteria

Descriptor	Marks
determines the simplified exponential term	1
determines number of litres	1

Q14c

2 marks

Use information from the table and the trapezoidal rule to determine the approximate volume of water in the vessel after three hours.

$t$	$\frac{dV}{dt}$
0	0.25
1	0.32
2	0.41
3	0.53

Reveal Answer

Using trapezoidal rule
Volume after 3 hours $= \frac{1}{2}(0.25+0.53+2(0.32+0.41))$

Volume after 3 hours $= 1.12$ litres

Marking Criteria

Descriptor	Marks
establishes expression for approximate number of litres of water in vessel after 3 hours	1
determines approximate number of litres	1

Q3

2020

QCAA

Paper 2

1 mark

Q3

1 mark

Let $R$ be the region enclosed by the graph of $y = xe^x$ , the $x$ -axis, and the lines $x = -1$ and $x = 1$ .

The area of $R$ is closest to

A

0.74

B

1.26

C

2.35

D

3.09

Reveal Answer

A

0.74

This is the value of the definite integral $\int_{-1}^{1} xe^x \, dx = \frac{2}{e} \approx 0.74$ . This calculation finds the net signed area, where the region below the x-axis cancels out part of the region above. To find the total geometric area, you must integrate the absolute value of the function.

B

1.26

Correct Answer

Since $y = xe^x$ is negative on $[-1, 0)$ and positive on $(0, 1]$ , the total area is calculated by splitting the integral: $\int_{-1}^{0} -(xe^x) \, dx + \int_{0}^{1} xe^x \, dx$ . Using integration by parts, this evaluates to $(1 - \frac{2}{e}) + 1 = 2 - \frac{2}{e} \approx 1.26$ .

C

2.35

This value corresponds to $\int_{-1}^{1} e^x \, dx = e - \frac{1}{e} \approx 2.35$ . This would be the area under the graph of $y=e^x$ , rather than the given function $y=xe^x$ .

D

3.09

This value approximates $|1 \cdot e^1| + |-1 \cdot e^{-1}| = e + \frac{1}{e} \approx 3.09$ . This calculation sums the absolute values of the function at the endpoints of the interval, which does not represent the area under the curve.

SCSA Mathematics Methods Integrals

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