| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Area under curve using integration |
| Difficulty | Moderate -0.3 This is a straightforward C2 integration question requiring algebraic manipulation (splitting the fraction), basic integration of power functions, and finding a definite integral. While it has multiple parts, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(\frac{1-8x^3}{x^2} = 0 \Rightarrow 1 - 8x^3 = 0\) | M1 |
| \(x^3 = \frac{1}{8}\) | M1 |
| \(x = \frac{1}{2}\) | A1 |
| Answer | Marks |
|---|---|
| \(= -x^{-1} - 4x^2 + c\) | M1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= -[-x^{-1} - 4x^2]_{\frac{1}{2}}^{2}\) | M1 | |
| \(= -\{(-\frac{1}{2} - 16) - (-2 - 1)\} = 13\frac{1}{2}\) | M1 A1 | (9) |
# Question 7:
## Part (i):
$\frac{1-8x^3}{x^2} = 0 \Rightarrow 1 - 8x^3 = 0$ | M1 |
$x^3 = \frac{1}{8}$ | M1 |
$x = \frac{1}{2}$ | A1 |
## Part (ii):
$f(x) = x^{-2} - 8x$
$\int f(x)\, dx = \int (x^{-2} - 8x)\, dx$
$= -x^{-1} - 4x^2 + c$ | M1 A2 |
## Part (iii):
$= -[-x^{-1} - 4x^2]_{\frac{1}{2}}^{2}$ | M1 |
$= -\{(-\frac{1}{2} - 16) - (-2 - 1)\} = 13\frac{1}{2}$ | M1 A1 | **(9)**
---7.\\
\includegraphics[max width=\textwidth, alt={}, center]{e5d62032-84ad-4e0b-9b72-ccfd8f4dbac8-3_499_721_248_552}
The diagram shows part of the curve $y = \mathrm { f } ( x )$ where $\mathrm { f } ( x ) = \frac { 1 - 8 x ^ { 3 } } { x ^ { 2 } } , x \neq 0$.\\
(i) Solve the equation $\mathrm { f } ( x ) = 0$.\\
(ii) Find $\int \mathrm { f } ( x ) \mathrm { d } x$.\\
(iii) Find the area of the shaded region bounded by the curve $y = \mathrm { f } ( x )$, the $x$-axis and the line $x = 2$.\\
\hfill \mbox{\textit{OCR C2 Q7 [9]}}