| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear guidance at each step. Part (i) is simple verification, part (ii) involves quotient rule differentiation (standard C3), and part (iii) provides the substitution explicitly with straightforward application. The final area calculation requires recognizing a geometric decomposition but is well-scaffolded. Slightly easier than average due to the extensive guidance provided. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x=3\), \(y = 3/\sqrt{(3-2)} = 3\), so P is \((3,3)\) which lies on \(y=x\) | M1 | Substituting \(x=3\) (both \(x\)'s) |
| A1 | \(y=3\) and completion ('\(3=3\)' is enough) | |
| [2] | or \(x = x/\sqrt{(x-2)}\) M1 \(\Rightarrow x=3\) A1 (by solving or verifying) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{\sqrt{x-2}\cdot 1 - x \cdot \frac{1}{2}(x-2)^{-1/2}}{x-2}\) | M1 | Quotient or product rule. PR: \(-\frac{1}{2}x(x-2)^{-3/2}+(x-2)^{-1/2}\) |
| A1 | Correct expression | |
| \(= \frac{x-2-\frac{1}{2}x}{(x-2)^{3/2}} = \frac{\frac{1}{2}x-2}{(x-2)^{3/2}}\) | M1 | \(\times\) top and bottom by \(\sqrt{x-2}\), e.g. taking out factor of \((x-2)^{-3/2}\) |
| \(= \frac{x-4}{2(x-2)^{3/2}}\) | A1 | NB AG |
| When \(x=3\), \(dy/dx = -\frac{1}{2}\times 1^{3/2} = -\frac{1}{2}\) | M1 | Substituting \(x=3\) |
| A1 | ||
| This gradient would be \(-1\) if curve were symmetrical about \(y=x\) | A1cao | Or an equivalent valid argument |
| [7] | If correct formula stated, allow one error; otherwise QR must be on correct \(u\) and \(v\), with numerator consistent with their derivatives and denominator correct initially. Allow ft on correct equivalent algebra from their incorrect expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = x-2 \Rightarrow du/dx = 1 \Rightarrow du = dx\) | B1 | or \(dx/du = 1\). No credit for integrating initial integral by parts. Condone \(du=1\). Condone missing \(du\)'s in subsequent working. |
| When \(x=3\), \(u=1\); when \(x=11\), \(u=9\) | B1 | |
| \(\Rightarrow \int_3^{11}\frac{x}{\sqrt{x-2}}\,dx = \int_1^9 \frac{u+2}{u^{1/2}}\,du\) | \(\int \frac{u+2}{u^{1/2}}\,(du)\) | |
| \(= \int_1^9 (u^{1/2}+2u^{-1/2})\,du\) | M1 | Splitting their fraction (correctly) and \(u/u^{1/2} = u^{1/2}\) (or \(\sqrt{u}\)) |
| \(= \left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]_1^9\) | A1 | \(\left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]\) (o.e.) |
| \(= (18+12)-(2/3+4)\) | M1 | Substituting correct limits |
| \(= 25\frac{1}{3}\) | A1cao | NB AG dep substitution and integration attempted |
| Area under \(y=x\) is \(\frac{1}{2}(3+11)\times 8 = 56\) | B1 | o.e. (e.g. \(60.5-4.5\)) |
| Area = (area under \(y=x\)) \(-\) (area under curve) | M1 | soi from working. Must be trapezium area: \(60.5 - 25\frac{1}{3}\) is M0 |
| So required area \(= 56 - 25\frac{1}{3} = 30\frac{2}{3}\) | A1cao | 30.7 or better |
| [9] |
# Question 2:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x=3$, $y = 3/\sqrt{(3-2)} = 3$, so P is $(3,3)$ which lies on $y=x$ | M1 | Substituting $x=3$ (both $x$'s) |
| | A1 | $y=3$ and completion ('$3=3$' is enough) |
| **[2]** | | or $x = x/\sqrt{(x-2)}$ M1 $\Rightarrow x=3$ A1 (by solving or verifying) |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{\sqrt{x-2}\cdot 1 - x \cdot \frac{1}{2}(x-2)^{-1/2}}{x-2}$ | M1 | Quotient or product rule. PR: $-\frac{1}{2}x(x-2)^{-3/2}+(x-2)^{-1/2}$ |
| | A1 | Correct expression |
| $= \frac{x-2-\frac{1}{2}x}{(x-2)^{3/2}} = \frac{\frac{1}{2}x-2}{(x-2)^{3/2}}$ | M1 | $\times$ top and bottom by $\sqrt{x-2}$, e.g. taking out factor of $(x-2)^{-3/2}$ |
| $= \frac{x-4}{2(x-2)^{3/2}}$ | A1 | **NB AG** |
| When $x=3$, $dy/dx = -\frac{1}{2}\times 1^{3/2} = -\frac{1}{2}$ | M1 | Substituting $x=3$ |
| | A1 | |
| This gradient would be $-1$ if curve were symmetrical about $y=x$ | A1cao | Or an equivalent valid argument |
| **[7]** | | If correct formula stated, allow one error; otherwise QR must be on correct $u$ and $v$, with numerator consistent with their derivatives and denominator correct initially. Allow ft on correct equivalent algebra from their incorrect expression |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = x-2 \Rightarrow du/dx = 1 \Rightarrow du = dx$ | B1 | or $dx/du = 1$. No credit for integrating initial integral by parts. Condone $du=1$. Condone missing $du$'s in subsequent working. |
| When $x=3$, $u=1$; when $x=11$, $u=9$ | B1 | |
| $\Rightarrow \int_3^{11}\frac{x}{\sqrt{x-2}}\,dx = \int_1^9 \frac{u+2}{u^{1/2}}\,du$ | | $\int \frac{u+2}{u^{1/2}}\,(du)$ |
| $= \int_1^9 (u^{1/2}+2u^{-1/2})\,du$ | M1 | Splitting their fraction (correctly) and $u/u^{1/2} = u^{1/2}$ (or $\sqrt{u}$) |
| $= \left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]_1^9$ | A1 | $\left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]$ (o.e.) |
| $= (18+12)-(2/3+4)$ | M1 | Substituting correct limits |
| $= 25\frac{1}{3}$ | A1cao | **NB AG** dep substitution and integration attempted |
| Area under $y=x$ is $\frac{1}{2}(3+11)\times 8 = 56$ | B1 | o.e. (e.g. $60.5-4.5$) |
| Area = (area under $y=x$) $-$ (area under curve) | M1 | soi from working. Must be trapezium area: $60.5 - 25\frac{1}{3}$ is M0 |
| So required area $= 56 - 25\frac{1}{3} = 30\frac{2}{3}$ | A1cao | 30.7 or better |
| **[9]** | | |
---2 Fig. 8 shows the curve $y = \frac { x } { \sqrt { x - 2 } }$, together with the lines $y = x$ and $x = 11$. The curve meets these lines at P and Q respectively. R is the point $( 11,11 )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0b4c4935-998c-404f-8fed-9b39b849168e-2_606_729_485_699}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Verify that the $x$-coordinate of P is 3 .\\
(ii) Show that, for the curve, $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }$.
Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about $y = x$.\\
(iii) Using the substitution $u = x - 2$, show that $\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }$.
Hence find the area of the region PQR bounded by the curve and the lines $y = x$ and $x = 11$.
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}