| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Find curve from gradient |
| Difficulty | Moderate -0.3 This is a straightforward C2 integration question with two standard parts: (i) integrating to find a curve equation using a boundary condition, and (ii) evaluating a definite integral with surds. Both require routine application of power rule integration and basic algebraic manipulation, making it slightly easier than average but still requiring careful execution. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks |
|---|---|
| \(y = \int \left(3 - \frac{2}{x}\right) dx\) | M1 A2 |
| \(y = 3x + 2x^{-1} + c\) | M1 |
| \((2, 6) \therefore 6 = 6 + 1 + c\) | A1 |
| \(c = -1\) | A1 |
| \(y = 3x + 2x^{-1} - 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_2^3 \left(6\sqrt{x} - \frac{4}{\sqrt{x}}\right) dx = \left[4x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\right]_2^3\) | M1 A2 | |
| \(= [4(3\sqrt{3}) - 8\sqrt{3}] - [4(2\sqrt{2}) - 8\sqrt{2}]\) | M1 B1 | |
| \(= (12\sqrt{3} - 8\sqrt{3}) - (8\sqrt{2} - 8\sqrt{2})\) | ||
| \(= 4\sqrt{3}\) \([k = 4]\) | A1 | (12) |
## Part (i)
$y = \int \left(3 - \frac{2}{x}\right) dx$ | M1 A2 |
$y = 3x + 2x^{-1} + c$ | M1 |
$(2, 6) \therefore 6 = 6 + 1 + c$ | A1 |
$c = -1$ | A1 |
$y = 3x + 2x^{-1} - 1$ | A1 |
## Part (ii)
$\int_2^3 \left(6\sqrt{x} - \frac{4}{\sqrt{x}}\right) dx = \left[4x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\right]_2^3$ | M1 A2 |
$= [4(3\sqrt{3}) - 8\sqrt{3}] - [4(2\sqrt{2}) - 8\sqrt{2}]$ | M1 B1 |
$= (12\sqrt{3} - 8\sqrt{3}) - (8\sqrt{2} - 8\sqrt{2})$ | |
$= 4\sqrt{3}$ $[k = 4]$ | A1 | **(12)**
---\begin{enumerate}[label=(\roman*)]
\item The gradient of a curve is given by
$$\frac{dy}{dx} = 3 - \frac{2}{x^2}, \quad x \neq 0.$$
Find an equation for the curve given that it passes through the point $(2, 6)$. [6]
\item Show that
$$\int_2^3 (6\sqrt{x} - \frac{4}{\sqrt{x}}) \, dx = k\sqrt{3},$$
where $k$ is an integer to be found. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR C2 Q8 [12]}}