QCAA Mathematical Methods Further integration

15 sample questions with marking guides and sample answers

Q5
2024
QCAA
Paper 1
1 mark
Q5
1 mark

Determine ∫ab2cos⁡(x)dx\int_a^b 2\cos(x)dx, where a=π3a = \frac{\pi}{3} and b=π2b = \frac{\pi}{2}

A

1−321 - \frac{\sqrt{3}}{2}

B

32−1\frac{\sqrt{3}}{2} - 1

C

2−32 - \sqrt{3}

D

3−2\sqrt{3} - 2

Reveal Answer
A

1−321 - \frac{\sqrt{3}}{2}

This option neglects the constant factor of 2. It represents the value of ∫π/3π/2cos⁡(x)dx\int_{\pi/3}^{\pi/2} \cos(x)dx, which is 1−321 - \frac{\sqrt{3}}{2}.

B

32−1\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

2−32 - \sqrt{3}

Correct Answer

The antiderivative of 2cos⁡(x)2\cos(x) is 2sin⁡(x)2\sin(x). Applying the Fundamental Theorem of Calculus yields 2sin⁡(π2)−2sin⁡(π3)=2(1)−2(32)=2−32\sin(\frac{\pi}{2}) - 2\sin(\frac{\pi}{3}) = 2(1) - 2(\frac{\sqrt{3}}{2}) = 2 - \sqrt{3}.

D

3−2\sqrt{3} - 2

This answer results from swapping the limits of integration or subtracting in the wrong order (F(a)−F(b)F(a) - F(b)), yielding the negative of the correct answer.

Q13
2021
QCAA
Paper 1
5 marks
Q13

Consider the functions f(x)=x2f(x)=x^2 and g(x)=4xg(x)=4x.

Q13a
2 marks

Determine the xx-coordinates of the points of intersection of the graphs of the two functions.

Reveal Answer

Solving simultaneously
x2=4xx^2 = 4x
Rearranging and factorising
x(x−4)=0x(x - 4) = 0
∴x=0\therefore x = 0 and x=4x = 4

Marking Criteria
DescriptorMarks

correctly uses the simultaneous procedure

1

correctly determines both x-intercept ordinates

1
Q13b
3 marks

Use the results from Question 13a) to calculate the area enclosed by the graphs of f(x)f(x) and g(x)g(x).

Reveal Answer

Area =∫04(4x−x2)dx= \int_0^4 (4x - x^2)dx
=[2x2−x33]04= \left[ 2x^2 - \frac{x^3}{3} \right]_0^4
=(2×42−433)−0= \left( 2 \times 4^2 - \frac{4^3}{3} \right) - 0

=323= \frac{32}{3} square units

Marking Criteria
DescriptorMarks

correctly determines the integral

1

substitutes limits into integral

1

determines area

1
Q6
2023
QCAA
Paper 2
1 mark
Q6
1 mark

∫a5a1x+adx\int_a^{5a} \frac{1}{x+a} dx, a≠0a \neq 0 is

A

1.7918

B

1.6094

C

1.3863

D

1.0986

Reveal Answer
A

1.7918

This value corresponds to ln⁡(6)\ln(6). This error typically occurs if you evaluate the antiderivative at the upper limit (x=5ax=5a) but forget to subtract the evaluation at the lower limit (x=ax=a).

B

1.6094

This value corresponds to ln⁡(5)\ln(5). This is incorrect; the integration results in the natural log of the ratio of the bounds adjusted by aa, which is ln⁡(3)\ln(3), not ln⁡(5)\ln(5).

C

1.3863

This value corresponds to ln⁡(4)\ln(4). The correct evaluation of the definite integral yields ln⁡(3)\ln(3).

D

1.0986

Correct Answer

The antiderivative is ln⁡∣x+a∣\ln|x+a|. Applying the limits yields ln⁡(5a+a)−ln⁡(a+a)=ln⁡(6a)−ln⁡(2a)=ln⁡(6a2a)=ln⁡(3)≈1.0986\ln(5a+a) - \ln(a+a) = \ln(6a) - \ln(2a) = \ln(\frac{6a}{2a}) = \ln(3) \approx 1.0986.

Q18
2025
QCAA
Paper 1
6 marks
Q18
6 marks

Two objects are launched simultaneously from different positions and travel along the same straight-line path. The objects are launched towards each other with the same initial speed.

The first object's displacement (m) from the origin is given by d1=13t3−12t2+ktd_1 = \frac{1}{3}t^3 - \frac{1}{2}t^2 + kt, where tt is the time (s) since the objects were launched and kk is a constant, k≠0k \neq 0. The second object is moving with a constant acceleration of 4 m s−24 \text{ m s}^{-2}.

The second object changes its direction, and at time t=1 st = 1 \text{ s} the objects have equal velocities and continue to travel in the same direction.

Compared to the first object, how much further does the second object travel between t=1 st = 1 \text{ s} and the next time the objects have equal velocities?

Reveal Answer

Finding velocity of the first object:

v1(t)=ddt(13t3−12t2+kt)=t2−t+kv_1(t) = \frac{d}{dt}(\frac{1}{3}t^3 - \frac{1}{2}t^2 + kt) = t^2 - t + k

Find the rule for the velocity for the second object:

v2(t)=∫4dt=4t+cv_2(t) = \int 4 dt = 4t + c

Given condition for

v1(1)=v2(1)12−1+k=4+cc=k−4\begin{align*} v_1(1) &= v_2(1)\\ 1^2 - 1 + k &= 4 + c\\ c &= k - 4 \end{align*}

Find the time when the velocities are the same.

v1(t)=v2(t)t2−t+k=4t+k−4t2−5t+4=0(t−1)(t−4)=0∴t=1,4\begin{align*} v_1(t) &= v_2(t)\\ t^2 - t + k &= 4t + k - 4\\ t^2 - 5t + 4 &= 0\\ (t - 1)(t - 4) &= 0\\ \therefore t &= 1, 4 \end{align*}

Need the time different to the given 1 s, so t=4t = 4.

As objects travel in the one direction between given times, the distance travelled is equal to the displacement. Find the difference between the objects' travelling distances.

displacement=∫velocity dtd2−d1=∫14(v2−v1)dt=∫14((4t−2)−(t2−t+2))dt=∫14(−t2+5t−4)dt=[−13t3+52t2−4t]14=−13×(4)3+52×(4)2−4×4−(−13+52−4)=−633+28−52=−21+28−2.5=4.5\begin{align*} \text{displacement} &= \int \text{velocity} \, dt\\ d_2 - d_1 &= \int_1^4 (v_2 - v_1) dt\\ &= \int_1^4 ((4t - 2) - (t^2 - t + 2)) dt\\ &= \int_1^4 (-t^2 + 5t - 4) dt\\ &= \left[ \frac{-1}{3}t^3 + \frac{5}{2}t^2 - 4t \right]_1^4\\ &= \frac{-1}{3} \times (4)^3 + \frac{5}{2} \times (4)^2 - 4 \times 4 - (\frac{-1}{3} + \frac{5}{2} - 4)\\ &= \frac{-63}{3} + 28 - \frac{5}{2}\\ &= -21 + 28 - 2.5\\ &= 4.5 \end{align*}
Marking Criteria
DescriptorMarks

correctly determines the equation for the first object's velocity

1

correctly determines the equation for the second object's velocity including the constant c

1

determine the relationship between constants kk and cc

1

determines the second time when the two velocities are the same

1

uses a suitable method for determining the difference between distances the objects have travelled in the given time interval

1

determines the difference between distances of the two objects

1
Q5
2023
VCAA
Paper 1
4 marks
Q5a
1 mark

Evaluate ∫0π3sin⁡(x)dx\int_0^{\frac{\pi}{3}} \sin(x) dx.

Reveal Answer

12\frac{1}{2}

Marking Criteria
DescriptorMarks

Evaluates the definite integral to find the correct answer of 12\frac{1}{2}

1
Q5b
3 marks

Hence, or otherwise, find all values of kk such that ∫0π3sin⁡(x)dx=∫kπ2cos⁡(x)dx\int_0^{\frac{\pi}{3}} \sin(x) dx = \int_k^{\frac{\pi}{2}} \cos(x) dx, where −3π<k<2π-3\pi < k < 2\pi.

Reveal Answer

∫kπ2cos⁡(x)dx=[sin⁡(x)]kπ2=sin⁡(π2)−sin⁡(k)=1−sin⁡(k)\int_k^{\frac{\pi}{2}} \cos(x)dx = [\sin(x)]_k^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(k) = 1 - \sin(k)

Using part a.,

12=1−sin⁡(k)\frac{1}{2} = 1 - \sin(k)

sin⁡(k)=12\sin(k) = \frac{1}{2}

k=−11π6,−7π6,π6,5π6k = \frac{-11\pi}{6}, \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}

Marking Criteria
DescriptorMarks

Correctly evaluates the right-hand side integral to obtain 1−sin⁡(k)1 - \sin(k)

1

Equates the evaluated integral to the answer from part a. and simplifies to sin⁡(k)=12\sin(k) = \frac{1}{2}

1

Finds all correct values of kk within the given domain: −11π6,−7π6,π6,5π6\frac{-11\pi}{6}, \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}

1
Q17
2022
QCAA
Paper 2
4 marks
Q17
4 marks

A snail is travelling along a straight path from point AA. The snail's velocity (cm min−1^{-1}) is modelled by v(t)=1.4ln⁡(1+t2)v(t)=1.4\ln\left(1+t^2\right), where tt is time (in minutes) for 0≤t≤150 \leq t \leq 15.

An ant passes point AA 12 minutes after the snail and follows the snail's path. The ant moves with a constant acceleration of 2 cm min−2^{-2} and passes the snail at t=15t=15 minutes.

Determine the ant's velocity at point AA.

Reveal Answer

Total displacement of the snail
∫0151.4ln⁡(1+t2)dt=76.0431\int_0^{15} 1.4 \ln(1 + t^2) dt = 76.0431 cm

Velocity of the ant =∫2dt= \int 2 dt
=2t+c= 2t + c

Displacementant from 12 to 15min=_{ant \ from \ 12 \ to \ 15min} = Displacementsnail from 0 to 15min_{snail \ from \ 0 \ to \ 15min}
∴∫12152t+c=76.0431\therefore \int_{12}^{15} 2t + c = 76.0431

Solving numerically on GDC
c=−1.6523c = -1.6523

Therefore, velocity of ant at t=12t = 12
=2×12−1.6523= 2 \times 12 - 1.6523
=22.3477= 22.3477 cm min−1^{-1} along the ant's path.

Marking Criteria
DescriptorMarks

correctly determines the total displacement of the snail

1

establishes an equation linking the ant and the snail

1

determines constant

1

determines velocity of ant

1
Q8
2022
VCAA
Paper 2
1 mark
Q8
1 mark

If ∫0bf(x)dx=10\int_0^b f(x) dx = 10 and ∫0af(x)dx=−4\int_0^a f(x) dx = -4, where 0<a<b0 < a < b, then ∫abf(x)dx\int_a^b f(x) dx is equal to

A

−6-6

B

−4-4

C

00

D

1010

E

1414

Reveal Answer
A

−6-6

This incorrectly subtracts 1010 from −4-4 or adds the two values incorrectly. The correct operation is to subtract the integral from 00 to aa from the integral from 00 to bb.

B

−4-4

This is simply the given value for ∫0af(x)dx\int_0^a f(x) dx, not the integral over the interval from aa to bb.

C

00

There is no mathematical justification for the integral evaluating to 00 based on the given values.

D

1010

This is the given value for ∫0bf(x)dx\int_0^b f(x) dx, which represents the area over the entire interval from 00 to bb, not just from aa to bb.

E

1414

Correct Answer

Using the additive property of definite integrals, ∫abf(x)dx=∫0bf(x)dx−∫0af(x)dx\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)=1410 - (-4) = 14.

Q3
2020
QCAA
Paper 2
1 mark
Q3
1 mark

Let RR be the region enclosed by the graph of y=xexy = xe^x, the xx-axis, and the lines x=−1x = -1 and x=1x = 1.

The area of RR 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 ∫−11xex dx=2e≈0.74\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=xexy = xe^x is negative on [−1,0)[-1, 0) and positive on (0,1](0, 1], the total area is calculated by splitting the integral: ∫−10−(xex) dx+∫01xex dx\int_{-1}^{0} -(xe^x) \, dx + \int_{0}^{1} xe^x \, dx. Using integration by parts, this evaluates to (1−2e)+1=2−2e≈1.26(1 - \frac{2}{e}) + 1 = 2 - \frac{2}{e} \approx 1.26.

C

2.35

This value corresponds to ∫−11ex dx=e−1e≈2.35\int_{-1}^{1} e^x \, dx = e - \frac{1}{e} \approx 2.35. This would be the area under the graph of y=exy=e^x, rather than the given function y=xexy=xe^x.

D

3.09

This value approximates ∣1⋅e1∣+∣−1⋅e−1∣=e+1e≈3.09|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.

Q11
2021
VCAA
Paper 2
1 mark
Q11
1 mark

If ∫0af(x)dx=k\int_0^a f(x)dx = k, then ∫0a(3f(x)+2)dx\int_0^a (3f(x) + 2)dx is

A

3k+2a3k + 2a

B

3k3k

C

k+2ak + 2a

D

k+2k + 2

E

3k+23k + 2

Reveal Answer
A

3k+2a3k + 2a

Correct Answer

Correct. By the linearity of integrals, ∫0a(3f(x)+2)dx=3∫0af(x)dx+∫0a2dx=3k+2a\int_0^a (3f(x) + 2)dx = 3\int_0^a f(x)dx + \int_0^a 2dx = 3k + 2a.

B

3k3k

Incorrect. This evaluates ∫0a3f(x)dx\int_0^a 3f(x)dx but completely ignores the integral of the constant 22.

C

k+2ak + 2a

Incorrect. This correctly evaluates the integral of 22 as 2a2a, but forgets to multiply the integral of f(x)f(x) by the constant factor 33.

D

k+2k + 2

Incorrect. This forgets to multiply the integral of f(x)f(x) by 33 and incorrectly evaluates ∫0a2dx\int_0^a 2dx as 22 instead of 2a2a.

E

3k+23k + 2

Incorrect. This correctly multiplies the integral of f(x)f(x) by 33, but incorrectly evaluates ∫0a2dx\int_0^a 2dx as 22 instead of 2a2a.

Q18
2024
QCAA
Paper 2
5 marks
Q18
5 marks

An object experiencing straight-line motion along a path has an acceleration (m s−2^{-2}) defined by the function a(t)=3sin⁡(2t)a(t) = 3\sin(2t) where tt is time (s) since the object begins moving, t≥0t \ge 0.

When t=0t = 0, both displacement and velocity are zero.

On the path is a motion sensor that is able to detect motion up to 2 metres away.

The object passes directly by the motion sensor when t=3t = 3.

Determine the average velocity of the object while it moves through the range of the sensor.

Reveal Answer

v(t)=∫a(t) dtv(t)=\int a(t)\,dt

=∫3sin⁡(2t) dt=\int 3\sin(2t)\,dt

=−32cos⁡(2t)+c=-\dfrac{3}{2}\cos(2t)+c

v(0)=0v(0)=0

0=−32cos⁡(0)+c0=-\dfrac{3}{2}\cos(0)+c

0=−32×1+c0=-\dfrac{3}{2}\times 1+c

c=32c=\dfrac{3}{2}

∴v(t)=−32cos⁡(2t)+32\therefore v(t)=-\dfrac{3}{2}\cos(2t)+\dfrac{3}{2}

d(t)=∫v(t) dtd(t)=\int v(t)\,dt

=∫[−32cos⁡(2t)+32]dt=\int\left[-\dfrac{3}{2}\cos(2t)+\dfrac{3}{2}\right]dt

=−34sin⁡(2t)+32t+c=-\dfrac{3}{4}\sin(2t)+\dfrac{3}{2}t+c

d(0)=0d(0)=0

c=0c=0

d(t)=−34sin⁡(2t)+32td(t)=-\dfrac{3}{4}\sin(2t)+\dfrac{3}{2}t

Using a GDC:

When t=3t=3,

d(3)=4.70956 md(3)=4.70956\ \text{m}

The sensor can detect motion 2 metres away, therefore
between d=2.70956d=2.70956 and d=6.70956d=6.70956

Find the times associated with these:

Using a GC;

d=2.70956d=2.70956

t=1.68914t=1.68914

d=6.70956d=6.70956

t=4.59214t=4.59214

The total time within sensor range is:

4.5921−1.68914.5921-1.6891

=2.90296 sec=2.90296\ \text{sec}

Average velocity within sensor range:

vˉ=42.90296\bar v=\dfrac{4}{2.90296}

=1.378 m/s=1.378\ \text{m/s}

Marking Criteria
DescriptorMarks

Correctly determines the velocity formula

1

Determines the displacement formula

1

Determines the object displacement when t=3t=3

1

Determines the time when the object is within sensor range

1

Determines the average velocity

1
Q1
2025
QCAA
Paper 1
1 mark
Q1
1 mark

Determine the value of ∫123x2 dx\int_{1}^{2} 3x^2 \, dx

A

9

B

7

C

6

D

5

Reveal Answer
A

9

Incorrect. This result likely comes from adding the evaluated bounds (23+13=92^3 + 1^3 = 9) instead of subtracting them, which violates the Fundamental Theorem of Calculus.

B

7

Correct Answer

Correct. The antiderivative of 3x23x^2 is x3x^3. Evaluating this from 1 to 2 using the Fundamental Theorem of Calculus gives 23−13=8−1=72^3 - 1^3 = 8 - 1 = 7.

C

6

Incorrect. This value might come from evaluating the derivative 6x6x at x=1x=1, rather than finding the definite integral.

D

5

Incorrect. This is a miscalculation. Remember to find the antiderivative x3x^3 and evaluate F(2)−F(1)F(2) - F(1).

Q7
2022
QCAA
Paper 2
1 mark
Q7
1 mark

A marble moves in one direction in a straight line with velocity v=2ln⁡(t+1)v = 2\ln(t + 1) (in metres per second) where tt is time (in seconds) since the marble passed through the origin.

Determine the distance from the origin the marble has rolled after 4 seconds.

A

0.40 m

B

1.60 m

C

3.22 m

D

8.09 m

Reveal Answer
A

0.40 m

This value represents the acceleration at t=4t=4, found by differentiating the velocity function: a(4)=v′(4)=24+1=0.40 m/s2a(4) = v'(4) = \frac{2}{4+1} = 0.40 \text{ m/s}^2.

B

1.60 m

This value corresponds to ln⁡(5)\ln(5), which is an intermediate value in the calculation but not the final integral result.

C

3.22 m

This is the instantaneous velocity of the marble at t=4t=4, calculated as v(4)=2ln⁡(5)≈3.22 m/sv(4) = 2\ln(5) \approx 3.22 \text{ m/s}, rather than the total distance traveled.

D

8.09 m

Correct Answer

Distance is found by integrating velocity from t=0t=0 to t=4t=4: ∫042ln⁡(t+1) dt=[2((t+1)ln⁡(t+1)−(t+1))]04=10ln⁡(5)−8≈8.09 m\int_0^4 2\ln(t+1) \, dt = [2((t+1)\ln(t+1) - (t+1))]_0^4 = 10\ln(5) - 8 \approx 8.09 \text{ m}.

Q5
2023
QCAA
Paper 1
1 mark
Q5
1 mark

Determine ∫1312xdx\int_1^3 \frac{1}{2x} dx.

A

12ln⁡6\frac{1}{2} \ln 6

B

12ln⁡5\frac{1}{2} \ln 5

C

12ln⁡4\frac{1}{2} \ln 4

D

12ln⁡3\frac{1}{2} \ln 3

Reveal Answer
A

12ln⁡6\frac{1}{2} \ln 6

This answer is incorrect. It corresponds to evaluating an antiderivative like 12ln⁡(2x)\frac{1}{2}\ln(2x) only at the upper limit x=3x=3, neglecting to subtract the value at the lower limit.

B

12ln⁡5\frac{1}{2} \ln 5

This is incorrect. The evaluation of the definite integral depends on the limits 11 and 33, which results in ln⁡(3)\ln(3), not ln⁡(5)\ln(5).

C

12ln⁡4\frac{1}{2} \ln 4

This option is incorrect. It may result from an arithmetic error or misapplying logarithmic properties, as the integral yields ln⁡(3)−ln⁡(1)\ln(3) - \ln(1), not ln⁡(4)\ln(4).

D

12ln⁡3\frac{1}{2} \ln 3

Correct Answer

We factor out the constant to get 12∫131xdx\frac{1}{2}\int_1^3 \frac{1}{x} dx. The antiderivative is 12[ln⁡∣x∣]13=12(ln⁡3−ln⁡1)\frac{1}{2}[\ln|x|]_1^3 = \frac{1}{2}(\ln 3 - \ln 1), which simplifies to 12ln⁡3\frac{1}{2}\ln 3 since ln⁡1=0\ln 1 = 0.

Q16
2025
QCAA
Paper 1
5 marks
Q16
5 marks

The energy produced by solar panels on a house is stored in a battery. Energy consumed in the house comes out of the battery.

The rate of energy produced by the solar panels during daylight hours on a particular day is approximately modelled by R(t)=2.5sin⁡(π12(t−6))R(t) = 2.5\sin(\frac{\pi}{12}(t - 6)), 6≤t≤186 \le t \le 18, where tt is the number of hours after midnight. R(t)=0R(t) = 0 for any hours after sunset and before sunrise.

The rate of energy consumed in the house is 4π\frac{4}{\pi} units per hour.

Determine the total energy change in the battery during daylight hours, assuming that sunrise is at 6:00 am and sunset is at 6:00 pm.

Reveal Answer

Total energy produced (units):

P(t)=∫618(2.5sin⁡(π12(t−6)))dt=[−30πcos⁡(π12(t−6))]618=−30πcos⁡(π12×12)+30πcos⁡(π12×0)=−30πcos⁡(π)+30πcos⁡(0)=30π+30π=60π\begin{align*} P(t) &= \int_6^{18} \left( 2.5\sin\left(\frac{\pi}{12}(t - 6)\right) \right) dt\\ &= \left[ -\frac{30}{\pi}\cos\left(\frac{\pi}{12}(t - 6)\right) \right]_6^{18}\\ &= -\frac{30}{\pi}\cos\left(\frac{\pi}{12} \times 12\right) + \frac{30}{\pi}\cos\left(\frac{\pi}{12} \times 0\right)\\ &= -\frac{30}{\pi}\cos(\pi) + \frac{30}{\pi}\cos(0)\\ &= \frac{30}{\pi} + \frac{30}{\pi}\\ &= \frac{60}{\pi} \end{align*}

Total energy consumed (units):

C(t)=12×4π=48πC(t) = 12 \times \frac{4}{\pi} = \frac{48}{\pi}

Total energy change:

P(t)−C(t)=60π−48π=12πP(t) - C(t) = \frac{60}{\pi} - \frac{48}{\pi} = \frac{12}{\pi}
Marking Criteria
DescriptorMarks

correctly integrates the energy produced term

1

determines the energy produced between sunrise and sunset

1

correctly determines a method to find the energy consumed between sunrise and sunset

1

determines the energy consumed during daylight hours

1

determines the total energy change between sunrise and sunset

1
Q5
2021
SCSA
Paper 1
6 marks
Q5a
4 marks

Determine the area between the parabola y=x2−x+3y = x^2 - x + 3 and the straight line y=x+3y = x + 3.

Reveal Answer

Point of intersection:

x2−x+3=x+3x2−2x=0x(x−2)=0∴x=0,2∫02[(x+3)−(x2−x+3)]dx=∫02[2x−x2]dx=[x2−x33]02=4−83=43 units2\begin{aligned} &x^2 - x + 3 = x + 3 \\ &x^2 - 2x = 0 \\ &x(x - 2) = 0 \\ &\therefore x = 0, 2 \\ &\int_0^2 \left[ (x + 3) - (x^2 - x + 3) \right] dx \\ &= \int_0^2 \left[ 2x - x^2 \right] dx \\ &= \left[ x^2 - \frac{x^3}{3} \right]_0^2 \\ &= 4 - \frac{8}{3} \\ &= \frac{4}{3} \text{ units}^2 \end{aligned}
Marking Criteria
DescriptorMarks

determines xx coordinates of the points of intersection

1

states correct integral for area

1

evaluates integral

1

determines correct area

1
Q5b
2 marks

The area between the parabola y=x2−x−2y = x^2 - x - 2 and the straight line y=x−2y = x - 2 is the same as the area determined in part (a). Explain why this is the case.

Reveal Answer

Both graphs from part (a) have been vertically translated down 5 units.
The shape of both graphs is unchanged.
Therefore, the area between them remains unchanged.

Marking Criteria
DescriptorMarks

states both graphs have been translated in the same direction by the same amount

1

states both graphs retain the same shape

1

Frequently Asked Questions

How many QCAA Mathematical Methods questions cover Further integration?
AusGrader has 151 QCAA Mathematical Methods questions on Further integration, all with instant AI grading and detailed marking feedback.

Ready to practise QCAA Mathematical Methods?

Get instant AI feedback on past exam questions, aligned to the syllabus

Start Practising Free