| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find curve equation from derivative (straightforward integration + point) |
| Difficulty | Easy -1.2 This is a straightforward integration question requiring only basic knowledge of standard integrals (polynomial and x^{-2}). The two-part structure is simple: integrate, then apply an initial condition to find the constant. No problem-solving insight needed, just routine application of C1 techniques. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 5x + \frac{x^{-1}}{-1} (+c)\) or \(y = 5x - x^{-1} + c\) | M1 A1 A1 | M1 for attempt to integrate (increase in power seen). A1 for \(5x\), A1 for \(-x^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Substituting \(y=7\), \(x=1\): \(7 = 5(1) - 1 + c \Rightarrow c = 3\) | M1 A1 | M1 for substituting values to find \(c\) |
| \(y = 5(2) - \frac{1}{2} + 3 = 12.5\) | M1 A1 | M1 for substituting \(x=2\) into their expression |
# Question 1:
## Part (a):
| $y = 5x + \frac{x^{-1}}{-1} (+c)$ or $y = 5x - x^{-1} + c$ | M1 A1 A1 | M1 for attempt to integrate (increase in power seen). A1 for $5x$, A1 for $-x^{-1}$ |
## Part (b):
| Substituting $y=7$, $x=1$: $7 = 5(1) - 1 + c \Rightarrow c = 3$ | M1 A1 | M1 for substituting values to find $c$ |
| $y = 5(2) - \frac{1}{2} + 3 = 12.5$ | M1 A1 | M1 for substituting $x=2$ into their expression |
---1.
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 5 + \frac { 1 } { x ^ { 2 } } .$$
\begin{enumerate}[label=(\alph*)]
\item Use integration to find $y$ in terms of $x$.
\item Given that $y = 7$ when $x = 1$, find the value of $y$ at $x = 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q1 [7]}}