| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Find indefinite integral of polynomial/power |
| Difficulty | Easy -1.2 This is a straightforward application of standard integration rules for powers of x. Each term integrates independently using the power rule, requiring only recall of the formula and basic algebraic manipulation. No problem-solving or conceptual insight needed—purely routine technique for a P1 question. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\left(\frac{8}{3}x^3 - \frac{1}{2}x^{-\frac{1}{2}} - 5\right)dx = \frac{8}{3} \times \frac{x^4}{4} - \frac{1}{2} \times 2x^{\frac{1}{2}} - 5x + c\) | M1 A1 | M1: Raising a power of \(x\) by 1 at least once, i.e. \(x^n \to x^{n+1}\) or \(-5 \to -5x\). Index does not have to be processed. Award following incorrect manipulations e.g. \(-\frac{1}{2\sqrt{x}} \to ...x^{\frac{1}{2}} \to ...x^{\frac{3}{2}}\) |
| A1 | For two of \(\frac{8}{3} \times \frac{x^4}{4}\), \(-\frac{1}{2} \times 2x^{\frac{1}{2}}\), \(-5x^1\) correct (unsimplified). May be seen on different lines. Indices must be processed. | |
| \(= \frac{2}{3}x^4 - x^{\frac{1}{2}} - 5x + c\) | A1 A1 | For two of \(\frac{2}{3}x^4\), \(-x^{\frac{1}{2}}\), \(-5x\) correct in simplest form. Accept \(\frac{2x^4}{3}\) and \(-\sqrt{x}\). Condone \(-5x^1\), \(-1x^{\frac{1}{2}}\) but not \(-\frac{5x}{1}\). Allow \(\frac{2}{3}\) as \(0.\dot{6}\) or \(0.6666...\) only. Final A1: fully correct on one line, simplified with \(+c\). |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\left(\frac{8}{3}x^3 - \frac{1}{2}x^{-\frac{1}{2}} - 5\right)dx = \frac{8}{3} \times \frac{x^4}{4} - \frac{1}{2} \times 2x^{\frac{1}{2}} - 5x + c$ | M1 A1 | M1: Raising a power of $x$ by 1 at least once, i.e. $x^n \to x^{n+1}$ or $-5 \to -5x$. Index does not have to be processed. Award following incorrect manipulations e.g. $-\frac{1}{2\sqrt{x}} \to ...x^{\frac{1}{2}} \to ...x^{\frac{3}{2}}$ |
| | A1 | For two of $\frac{8}{3} \times \frac{x^4}{4}$, $-\frac{1}{2} \times 2x^{\frac{1}{2}}$, $-5x^1$ correct (unsimplified). May be seen on different lines. Indices must be processed. |
| $= \frac{2}{3}x^4 - x^{\frac{1}{2}} - 5x + c$ | A1 A1 | For two of $\frac{2}{3}x^4$, $-x^{\frac{1}{2}}$, $-5x$ correct in simplest form. Accept $\frac{2x^4}{3}$ and $-\sqrt{x}$. Condone $-5x^1$, $-1x^{\frac{1}{2}}$ but not $-\frac{5x}{1}$. Allow $\frac{2}{3}$ as $0.\dot{6}$ or $0.6666...$ only. Final A1: fully correct **on one line**, simplified with $+c$. |
---\begin{enumerate}
\item Find, in simplest form,
\end{enumerate}
$$\int \left( \frac { 8 x ^ { 3 } } { 3 } - \frac { 1 } { 2 \sqrt { x } } - 5 \right) \mathrm { d } x$$
\hfill \mbox{\textit{Edexcel P1 2020 Q1 [4]}}