| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| 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 Part (i) requires rewriting the integrand in index form and applying standard power rule integration (routine algebraic manipulation), while part (ii) is a straightforward reverse problem involving basic polynomial integration and solving a linear equation. Both parts are standard textbook exercises requiring only direct application of techniques with no problem-solving insight, making this slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{8\sqrt{x}-5}{2x^2} \rightarrow 4x^{-\frac{3}{2}} - \frac{5}{2}x^{-2}\) | B1 | Seen or implied. Ignore any \(+C\). May also see factor of ½ taken out. |
| \(= -8x^{-\frac{1}{2}} + \frac{5}{2}x^{-1}\ (+C)\) | M1 A1 | M1: raising any index by one on one term. A1: correct terms (indices must be processed) |
| \(\int_2^4\left(4x^{-\frac{3}{2}}-\frac{5}{2}x^{-2}\right)dx = \left(-4+\frac{5}{8}\right)-\left(-4\sqrt{2}+\frac{5}{4}\right) = 4\sqrt{2}-\frac{37}{8}\) | dM1 A1 | dM1: substitutes 4 and 2 and subtracts either way. A1: \(4\sqrt{2}-\frac{37}{8}\) or exact equivalent e.g. \(4\sqrt{2}-4.625\) or \(\frac{32\sqrt{2}-37}{8}\) |
| (5) | Note: Answer only scores 0 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\left(\frac{1}{2}x^2+k\right)dx = \left[\frac{1}{6}x^3+kx\right]\) | M1 A1 | M1: one term correct unsimplified e.g. \(\frac{\frac{1}{2}x^3}{3}\) or \(kx^1\), index must be processed. A1: \(\frac{1}{6}x^3+kx\) |
| \(\int_{-3}^6\left(\frac{1}{2}x^2+k\right)dx=55 \Rightarrow \left[\frac{1}{6}x^3+kx\right]_{-3}^6=55\) | dM1 | Substitutes 6 and \(-3\), subtracts, sets \(=55\), proceeds to \(k=\ldots\) |
| \(\Rightarrow (36+6k)-\left(-\frac{9}{2}-3k\right)=55\) | ||
| \(k = \frac{29}{18}\) | A1 | Or exact equivalent e.g. \(1.6\dot{1}\). Do not allow 1.6 or 1.61 or 1.61… |
| (4) |
## Question 8:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{8\sqrt{x}-5}{2x^2} \rightarrow 4x^{-\frac{3}{2}} - \frac{5}{2}x^{-2}$ | B1 | Seen or implied. Ignore any $+C$. May also see factor of ½ taken out. |
| $= -8x^{-\frac{1}{2}} + \frac{5}{2}x^{-1}\ (+C)$ | M1 A1 | M1: raising any index by one on one term. A1: correct terms (indices must be processed) |
| $\int_2^4\left(4x^{-\frac{3}{2}}-\frac{5}{2}x^{-2}\right)dx = \left(-4+\frac{5}{8}\right)-\left(-4\sqrt{2}+\frac{5}{4}\right) = 4\sqrt{2}-\frac{37}{8}$ | dM1 A1 | dM1: substitutes 4 and 2 and subtracts either way. A1: $4\sqrt{2}-\frac{37}{8}$ or exact equivalent e.g. $4\sqrt{2}-4.625$ or $\frac{32\sqrt{2}-37}{8}$ |
| | **(5)** | **Note: Answer only scores 0 marks** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{1}{2}x^2+k\right)dx = \left[\frac{1}{6}x^3+kx\right]$ | M1 A1 | M1: one term correct unsimplified e.g. $\frac{\frac{1}{2}x^3}{3}$ or $kx^1$, index must be processed. A1: $\frac{1}{6}x^3+kx$ |
| $\int_{-3}^6\left(\frac{1}{2}x^2+k\right)dx=55 \Rightarrow \left[\frac{1}{6}x^3+kx\right]_{-3}^6=55$ | dM1 | Substitutes 6 and $-3$, subtracts, sets $=55$, proceeds to $k=\ldots$ |
| $\Rightarrow (36+6k)-\left(-\frac{9}{2}-3k\right)=55$ | | |
| $k = \frac{29}{18}$ | A1 | Or exact equivalent e.g. $1.6\dot{1}$. Do not allow 1.6 or 1.61 or 1.61… |
| | **(4)** | |
---8. Solutions relying on calculator technology are not acceptable in this question.\\
(i)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bfeb1724-9a00-4a36-9606-520395792b2b-22_556_822_351_561}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of a curve with equation
$$y = \frac { 8 \sqrt { x } - 5 } { 2 x ^ { 2 } } \quad x > 0$$
The region $R$, shown shaded in Figure 2, is bounded by the curve, the line with equation $x = 2$, the $x$-axis and the line with equation $x = 4$
Find the exact area of $R$.\\
(ii) Find the value of the constant $k$ such that
$$\int _ { - 3 } ^ { 6 } \left( \frac { 1 } { 2 } x ^ { 2 } + k \right) \mathrm { d } x = 55$$
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel P2 2019 Q8 [9]}}