| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area Between Curve and Both Axes |
| Difficulty | Moderate -0.3 This is a straightforward AS-level integration question testing understanding of area below the x-axis. Part (a) requires factorizing a simple quadratic. Part (b)(i) involves computing a definite integral and taking absolute values for regions below the axis—a standard textbook exercise. Part (b)(ii) tests conceptual understanding that integration gives signed area. While it requires careful attention to negative areas, it involves no novel problem-solving or advanced techniques. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) is \((0,2)\), \(B\) is \((1,0)\), \(C\) is \((2,0)\) | M1 | States correct coordinates of \(B\) or \(C\), or correct \(x\)-coordinates of \(B\) and \(C\) |
| \(A(0,2)\), \(B(1,0)\), \(C(2,0)\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 (x^2 - 3x + 2)\,dx\) | M1 | Integrates in two parts with limits \(O\) to \(B\) and \(B\) to \(C\) |
| \(= \left[\frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x\right]_0^1 = \frac{1}{3} - \frac{3}{2} + 2 = \frac{5}{6}\) | M1, A1 | Integrates quadratic with at least one term correct; integrates completely correctly |
| \(\int_1^2 (x^2 - 3x + 2)\,dx = \left[\frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x\right]_1^2\) | M1 | Substitutes two sets of limits and subtracts; takes modulus of \(BC\) value |
| \(= \left[\frac{1}{3}\times 8 - \frac{3}{2}\times 4 + 2\times 2\right] - \frac{5}{6} = -\frac{1}{6}\) | — | — |
| Total area \(= \frac{5}{6} - \left(-\frac{1}{6}\right) = 1\) | R1 | Completes calculation convincingly; AG |
| Answer | Marks | Guidance |
|---|---|---|
| The calculator treats the area between \(B\) and \(C\) as negative. | E1 | — |
## Question 7(a):
$A$ is $(0,2)$, $B$ is $(1,0)$, $C$ is $(2,0)$ | M1 | States correct coordinates of $B$ or $C$, or correct $x$-coordinates of $B$ and $C$ |
$A(0,2)$, $B(1,0)$, $C(2,0)$ | A1 | — |
## Question 7(b)(i):
$\int_0^1 (x^2 - 3x + 2)\,dx$ | M1 | Integrates in two parts with limits $O$ to $B$ and $B$ to $C$ |
$= \left[\frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x\right]_0^1 = \frac{1}{3} - \frac{3}{2} + 2 = \frac{5}{6}$ | M1, A1 | Integrates quadratic with at least one term correct; integrates completely correctly |
$\int_1^2 (x^2 - 3x + 2)\,dx = \left[\frac{1}{3}x^3 - \frac{3}{2}x^2 + 2x\right]_1^2$ | M1 | Substitutes two sets of limits and subtracts; takes modulus of $BC$ value |
$= \left[\frac{1}{3}\times 8 - \frac{3}{2}\times 4 + 2\times 2\right] - \frac{5}{6} = -\frac{1}{6}$ | — | — |
Total area $= \frac{5}{6} - \left(-\frac{1}{6}\right) = 1$ | R1 | Completes calculation convincingly; AG |
## Question 7(b)(ii):
The calculator treats the area between $B$ and $C$ as negative. | E1 | — |
---7 The diagram below shows the graph of the curve that has equation $y = x ^ { 2 } - 3 x + 2$ along with two shaded regions.\\
\includegraphics[max width=\textwidth, alt={}, center]{f87d1b36-26db-4a0b-b9ec-d7d82a396aba-08_646_711_408_667}
7
\begin{enumerate}[label=(\alph*)]
\item State the coordinates of the points $A , B$ and $C$.\\
7
\item Katy is asked by her teacher to find the total area of the two shaded regions.\\
Katy uses her calculator to find $\int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 3 x + 2 \right) \mathrm { d } x$ and gets the answer $\frac { 2 } { 3 }$\\
Katy's teacher says that her answer is incorrect.\\
7 (b) (i) Show that the total area of the two shaded regions is 1\\
Fully justify your answer.\\
7 (b) (ii) Explain why Katy's method was not valid.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2021 Q7 [8]}}