| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Moderate -0.8 Part (a) is a straightforward application of the trapezium rule with clearly defined stripsβa routine numerical methods question requiring only substitution into a formula. Part (b) asks for an explanation of method (integration of a quadratic) rather than execution, which is even more basic. Both parts are standard textbook exercises with no problem-solving or insight required, making this easier than average. |
| Spec | 1.09f Trapezium rule: numerical integration3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks |
|---|---|
| 15(a) | States strip width = 20 |
| Answer | Marks | Guidance |
|---|---|---|
| contradicted. β | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept all y values | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| far) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7970m and less than 8250m) | 1.1a | A1 |
| Subtotal | 4 |
| Answer | Marks |
|---|---|
| 15(b) | Explains means of gaining a more |
| Answer | Marks | Guidance |
|---|---|---|
| method. | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question Total | 5 | |
| Q | Marking Instructions | AO |
Question 15:
--- 15(a) ---
15(a) | States strip width = 20
PI by correct y values and not
contradicted. β | 1.1b | B1 | β = , 20 ,
, ,
π¦π¦0 = 131 π¦π¦1 = 140
π¦π¦2 = 120 π¦π¦3 = 80 π¦π¦4 = 0
β
π΄π΄π΄π΄ππππ = (π¦π¦0+2π¦π¦1+2π¦π¦2+2π¦π¦3
2
Distance+ =π¦π¦ 8 4 1)10 m
States five values PI
, ,
π¦π¦ , ,
π¦π¦0 = 131 π¦π¦1 = 140
If theπ¦π¦y 2 u=se1 2h0=25π¦π¦ 3 co=nd8o0neπ¦π¦ 4 = 0
y =6, y =135, y =134,
0 1 2
y =94, y =0
3 4
If they use five strips between
20<t<100 condone
y =131, y =140, y =132,
0 1 2
y =108, y =67,y =0
3 4 5
Β±2on
Accept all y values | 1.1a | M1
Applies correct trapezium rule
formula to their values (this mark
could be achieved with B0M0 so
far) | 1.1a | M1
Applies trapezium rule with four
strips and obtains correct value for
distance. (accept values between
7970m and less than 8250m) | 1.1a | A1
Subtotal | 4
--- 15(b) ---
15(b) | Explains means of gaining a more
accurate estimate
For example, βintegrate the
quadratic between 20 and 100β
Must include limits for integration.
Or
Use the quadratic to calculate y
values in the trapezium rule or
other appropriate numerical
method. | 2.4 | E1 | Integrates the quadratic between
the limits 20 and 100.
Subtotal | 1
Question Total | 5
Q | Marking Instructions | AO | Marks | Typical SolutionA particle is moving in a straight line with velocity $v\text{ ms}^{-1}$ at time $t$ seconds as shown by the graph below.
\includegraphics{figure_15}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with four strips to estimate the distance travelled by the particle during the time period $20 \leq t \leq 100$
[4 marks]
\item Over the same time period, the curve can be very closely modelled by a particular quadratic.
Explain how you could find an alternative estimate using this quadratic.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2020 Q15 [5]}}