| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (straightforward integration + point) |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring splitting the fraction, integrating term-by-term (x^3/x^3 - 4/x^3 = 1 - 4x^{-3}), finding the constant using the boundary condition, then substituting x=2. It's slightly easier than average as it involves routine algebraic manipulation and standard integration rules with no conceptual challenges, though the negative power and constant determination add minor complexity beyond pure recall. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08b Integrate x^n: where n != -1 and sums |
\begin{enumerate}
\item Given that
\end{enumerate}
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 3 } - 4 } { x ^ { 3 } } , \quad x \neq 0$$
and that $y = 0$ when $x = - 1$, find the value of $y$ when $x = 2$.\\
\hfill \mbox{\textit{OCR C2 Q6 [8]}}