| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Area under curve using substitution |
| Difficulty | Standard +0.3 This is a straightforward P2 integration question requiring expansion of the numerator, term-by-term integration with power rule, and substitution into limits. Part (b) uses simple coordinate substitution, and part (c) combines standard area-between-curves technique. All steps are routine textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{(x-k)^2}{\sqrt{x}} = \frac{x^2 - 2kx + k^2}{x^{\frac{1}{2}}} = x^{\frac{3}{2}} - 2kx^{\frac{1}{2}} + k^2x^{-\frac{1}{2}}\) | M1 | Multiply out numerator and split into separate terms; score for one term with correct processed index. Condone \((x-k)^2 \Rightarrow x^2 \pm k^2\) |
| \(\frac{2}{5}x^{\frac{5}{2}} - \frac{4}{3}kx^{\frac{3}{2}} + 2k^2x^{\frac{1}{2}}\) | M1, A1 | M1: raise power by one on at least one term (index must be fraction/decimal). A1: fully correct, indices processed, ignore spurious notation, \(+c\) optional |
| Substitute \(x=16\) and \(x=1\), subtract: \(\frac{2}{5}(16)^{\frac{5}{2}} - \frac{4}{3}k(16)^{\frac{3}{2}} + 2k^2(16)^{\frac{1}{2}} - \frac{2}{5} + \frac{4}{3}k - 2k^2\) | dM1 | Dependent on previous M1; substitute 16 and 1, subtract either way |
| \(= 6k^2 - 84k + \frac{2046}{5}\) | A1 | Accept e.g. \(6k^2 - 84k + 409.2\); withhold if spurious \(\int\) or \(dx\) remains |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9 = \frac{(1-k)^2}{\sqrt{1}} \Rightarrow 9 = (1-k)^2\) | M1 | Substitute coordinates of \(A\) into equation for \(C\), or substitute \(A\) and \(k=4\) |
| \(k = 4\) only | A1* | Must reject \(k=-2\) or clearly select \(k=4\); \(k^2 - 2k - 8 = 0\) acceptable route; minimal conclusion required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(q = 36\) | B1 | May appear in (b) or on diagram; implied by correct trapezium/triangle area |
| \(\frac{1}{2}(9 + 36) \times 15 - \left(6(4)^2 - 84(4) + \frac{2046}{5}\right) = \frac{1683}{10}\) | M1, A1 | M1: attempt area of \(R\) using their \(q\) and part (a) answer; condone arithmetic slips and wrong-way subtraction. A1: \(\frac{1683}{10}\) oe, ignore units, condone spurious notation |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(x-k)^2}{\sqrt{x}} = \frac{x^2 - 2kx + k^2}{x^{\frac{1}{2}}} = x^{\frac{3}{2}} - 2kx^{\frac{1}{2}} + k^2x^{-\frac{1}{2}}$ | M1 | Multiply out numerator and split into separate terms; score for one term with correct processed index. Condone $(x-k)^2 \Rightarrow x^2 \pm k^2$ |
| $\frac{2}{5}x^{\frac{5}{2}} - \frac{4}{3}kx^{\frac{3}{2}} + 2k^2x^{\frac{1}{2}}$ | M1, A1 | M1: raise power by one on at least one term (index must be fraction/decimal). A1: fully correct, indices processed, ignore spurious notation, $+c$ optional |
| Substitute $x=16$ and $x=1$, subtract: $\frac{2}{5}(16)^{\frac{5}{2}} - \frac{4}{3}k(16)^{\frac{3}{2}} + 2k^2(16)^{\frac{1}{2}} - \frac{2}{5} + \frac{4}{3}k - 2k^2$ | dM1 | Dependent on previous M1; substitute 16 and 1, subtract either way |
| $= 6k^2 - 84k + \frac{2046}{5}$ | A1 | Accept e.g. $6k^2 - 84k + 409.2$; withhold if spurious $\int$ or $dx$ remains |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9 = \frac{(1-k)^2}{\sqrt{1}} \Rightarrow 9 = (1-k)^2$ | M1 | Substitute coordinates of $A$ into equation for $C$, or substitute $A$ and $k=4$ |
| $k = 4$ only | A1* | Must reject $k=-2$ or clearly select $k=4$; $k^2 - 2k - 8 = 0$ acceptable route; minimal conclusion required |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $q = 36$ | B1 | May appear in (b) or on diagram; implied by correct trapezium/triangle area |
| $\frac{1}{2}(9 + 36) \times 15 - \left(6(4)^2 - 84(4) + \frac{2046}{5}\right) = \frac{1683}{10}$ | M1, A1 | M1: attempt area of $R$ using their $q$ and part (a) answer; condone arithmetic slips and wrong-way subtraction. A1: $\frac{1683}{10}$ oe, ignore units, condone spurious notation |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \quad x > 0$$
where $k$ is a positive constant.\\
(a) Show that
$$\int _ { 1 } ^ { 16 } \frac { ( x - k ) ^ { 2 } } { \sqrt { x } } \mathrm {~d} x = a k ^ { 2 } + b k + \frac { 2046 } { 5 }$$
where $a$ and $b$ are integers to be found.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0e3b364c-151b-471d-acb6-01afb018fb75-26_645_670_904_699}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ and the line $l$.\\
Given that $l$ intersects $C$ at the point $A ( 1,9 )$ and at the point $B ( 16 , q )$ where $q$ is a constant,\\
(b) show that $k = 4$
The region $R$, shown shaded in Figure 1, is bounded by $C$ and $l$\\
Using the answers to parts (a) and (b),\\
(c) find the area of region $R$
\hfill \mbox{\textit{Edexcel P2 2023 Q10 [10]}}