Easy -1.8 This is a straightforward multiple-choice question requiring recognition that the integral of x³ from -4 to 4 gives zero due to symmetry, so the total shaded area requires taking absolute values of the negative and positive regions. The calculation is simple: 2|∫₀⁴ x³ dx| = 2|64/4| = 128, with no problem-solving or novel insight needed.
Question 3:
3 | Circles correct answer | AO1.1b | B1 | 68
Total | 1
Q | Marking Instructions | AO | Marks | Typical Solution
4a | Sketches graph recalling correct u
shape
Deduces correct relative positions
of intersections with axes and with
k labelled | AO1.2 | B1 | k
AO2.2a | B1
b | Shows evidence of discriminant
being used or completing the
square to find vertex | AO1.1a | M1 | For distinct roots
b2 4ac0
62
41k 0
364k 0
k 9
k is the y-intercept and for positive
roots the intercept must be positive
0k 9
Obtains k 9
Condone k 9 | AO1.1b | A1
Explains that positive roots and the
u shape of the graph (OE) mean
the graph must cross the y-axis
above 0 or k > 0. | AO2.4 | E1
States correct range of values for k | AO2.2a | R1
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution
The graph of $y = x^3$ is shown.
\includegraphics{figure_1}
Find the total shaded area.
Circle your answer.
[1 mark]
$-68$ 60 68 128
\hfill \mbox{\textit{AQA Paper 2 2018 Q3 [1]}}