| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Combined region areas |
| Difficulty | Moderate -0.3 This is a straightforward C2 integration question requiring integration of a polynomial, finding roots of a cubic, and calculating areas. While it has multiple parts and requires careful handling of areas below the x-axis, all techniques are standard and the algebraic manipulation is routine for this level. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(y = \int(4 - 6x - 3x^2)\ dx\) | |
| \(y = 4x - 3x^2 - x^3 + c\) | M1 A2 |
| \((0,0)\ \therefore\ c = 0\) | M1 |
| \(y = 4x - 3x^2 - x^3\) | A1 |
| Answer | Marks |
|---|---|
| \(4x - 3x^2 - x^3 = 0\), \(-x(x+4)(x-1) = 0\) | M1 |
| \(x = 0\) (at \(O\)), \(-4\), \(1\) | |
| \(\therefore\ A(-4, 0),\ B(1, 0)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= -\int_{-4}^{0}(4x - 3x^2 - x^3)\ dx + \int_0^1(4x - 3x^2 - x^3)\ dx\) | M1 | |
| \(= -[2x^2 - x^3 - \frac{1}{4}x^4]_{-4}^{0} + [2x^2 - x^3 - \frac{1}{4}x^4]_0^1\) | M1 A1 | |
| \(= -[(0) - (32 + 64 - 64)] + [(2 - 1 - \frac{1}{4}) - (0)]\) | M1 | |
| \(= 32 + \frac{3}{4} = 32\frac{3}{4}\) | A1 | (12) |
# Question 9:
## Part (i):
$y = \int(4 - 6x - 3x^2)\ dx$ | |
$y = 4x - 3x^2 - x^3 + c$ | M1 A2 |
$(0,0)\ \therefore\ c = 0$ | M1 |
$y = 4x - 3x^2 - x^3$ | A1 |
## Part (ii):
$4x - 3x^2 - x^3 = 0$, $-x(x+4)(x-1) = 0$ | M1 |
$x = 0$ (at $O$), $-4$, $1$ | |
$\therefore\ A(-4, 0),\ B(1, 0)$ | A1 |
## Part (iii):
$= -\int_{-4}^{0}(4x - 3x^2 - x^3)\ dx + \int_0^1(4x - 3x^2 - x^3)\ dx$ | M1 |
$= -[2x^2 - x^3 - \frac{1}{4}x^4]_{-4}^{0} + [2x^2 - x^3 - \frac{1}{4}x^4]_0^1$ | M1 A1 |
$= -[(0) - (32 + 64 - 64)] + [(2 - 1 - \frac{1}{4}) - (0)]$ | M1 |
$= 32 + \frac{3}{4} = 32\frac{3}{4}$ | A1 | **(12)**9.\\
\includegraphics[max width=\textwidth, alt={}, center]{e4afa57d-5be3-42a6-ab35-39b0fdcc1681-3_559_732_824_388}
The diagram shows the curve with equation $y = \mathrm { f } ( x )$ which crosses the $x$-axis at the origin and at the points $A$ and $B$.
Given that
$$f ^ { \prime } ( x ) = 4 - 6 x - 3 x ^ { 2 }$$
(i) find an expression for $y$ in terms of $x$,\\
(ii) show that $A$ has coordinates ( $- 4,0$ ) and find the coordinates of $B$,\\
(iii) find the total area of the two regions bounded by the curve and the $x$-axis.
\hfill \mbox{\textit{OCR C2 Q9 [12]}}