| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line-Axis Bounded Region |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring finding intersection points by solving a quadratic equation, then calculating area between curves using definite integration. The algebra is straightforward (equating line and curve), and the integration involves basic polynomial integration with careful attention to which function is on top. Slightly above average difficulty due to the multi-step nature and need to split the region appropriately, but still a routine textbook exercise. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x-3 = x^2-2x-15\) so \(x^2-4x-12=0\) | M1 | Puts equations equal |
| \(x=6\) or \(x=-2\) | dM1 A1 | Solves quadratic; both answers correct |
| \(y=9\) or \(y=-7\) | dM1 A1 | Finds \(y\); both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int x^2-2x-15\,dx = \frac{1}{3}x^3 - x^2 - 15x\) | B1 | Correct integration of quadratic |
| Line meets \(x\)-axis at \(x=1\frac{1}{2}\); curve meets axis at \(x=5\) | B1 B1 | Line intersection correct (1.5); curve intersection correct (5); their 1.5 must NOT be zero |
| Uses correct combination of correct areas: Area = Area of large triangle MINUS \(\left[\frac{1}{3}x^3 - x^2 - 15x\right]_5^6\) | M1 | Uses correct combination; allow numerical slips |
| Area of large triangle \(= \frac{1}{2}\times(6-1\frac{1}{2})\times 9\) | dM1 | Attempts second area (triangle relevant to method) |
| \(= \frac{1}{2}(6-1\frac{1}{2})\times9 - \left[(\frac{1}{3}(6)^3-(6)^2-15(6))-(\frac{1}{3}(5)^3-(5)^2-15(5))\right]\) | M1 | Uses limits correctly on cubic |
| \(= 20.25 - (-54-(-58\frac{1}{3})) = \frac{191}{12} = 15\frac{11}{12}\) | A1 | Final answer — not decimal — cso |
## Question 14:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x-3 = x^2-2x-15$ so $x^2-4x-12=0$ | M1 | Puts equations equal |
| $x=6$ or $x=-2$ | dM1 A1 | Solves quadratic; both answers correct |
| $y=9$ or $y=-7$ | dM1 A1 | Finds $y$; both correct |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x^2-2x-15\,dx = \frac{1}{3}x^3 - x^2 - 15x$ | B1 | Correct integration of quadratic |
| Line meets $x$-axis at $x=1\frac{1}{2}$; curve meets axis at $x=5$ | B1 B1 | Line intersection correct (1.5); curve intersection correct (5); their 1.5 must NOT be zero |
| Uses correct combination of correct areas: Area = Area of large triangle MINUS $\left[\frac{1}{3}x^3 - x^2 - 15x\right]_5^6$ | M1 | Uses correct combination; allow numerical slips |
| Area of large triangle $= \frac{1}{2}\times(6-1\frac{1}{2})\times 9$ | dM1 | Attempts second area (triangle relevant to method) |
| $= \frac{1}{2}(6-1\frac{1}{2})\times9 - \left[(\frac{1}{3}(6)^3-(6)^2-15(6))-(\frac{1}{3}(5)^3-(5)^2-15(5))\right]$ | M1 | Uses limits correctly on cubic |
| $= 20.25 - (-54-(-58\frac{1}{3})) = \frac{191}{12} = 15\frac{11}{12}$ | A1 | Final answer — not decimal — cso |
---14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e878227b-d625-4ef2-ac49-a9dc05c5321a-36_652_791_223_548}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Diagram NOT drawn to scale
Figure 2 shows part of the line $l$ with equation $y = 2 x - 3$ and part of the curve $C$ with equation $y = x ^ { 2 } - 2 x - 15$
The line $l$ and the curve $C$ intersect at the points $A$ and $B$ as shown.
\begin{enumerate}[label=(\alph*)]
\item Use algebra to find the coordinates of $A$ and the coordinates of $B$.
In Figure 2, the shaded region $R$ is bounded by the line $l$, the curve $C$ and the positive $x$-axis.
\item Use integration to calculate an exact value for the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q14 [12]}}