| 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 verification by substitution, part (ii) uses quotient rule differentiation (standard C3), and part (iii) applies a given substitution u=x-2 to evaluate a definite integral. The final area calculation requires subtracting a triangle from the integral result. All techniques are routine for C3 level with no novel problem-solving required beyond following the scaffolded steps. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08f Area between two curves: using integration1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x=3\), \(y = 3/\sqrt{(3-2)} = 3\) | M1 | Substituting \(x=3\) (both \(x\)'s). Or \(x = x/\sqrt{(x-2)}\) M1 \(\Rightarrow x=3\) A1 (by solving or verifying) |
| So P is \((3,3)\) which lies on \(y=x\) | A1 | \(y=3\) and completion ('\(3=3\)' is enough) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{\sqrt{x-2}\cdot1 - x\cdot\frac{1}{2}(x-2)^{-1/2}}{x-2}\) | M1 | Quotient or product rule. 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 |
| 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 \(\sqrt{(x-2)}\) o.e. e.g. taking out factor of \((x-2)^{-3/2}\). Allow ft on correct equivalent algebra from their incorrect expression |
| \(= \frac{x-4}{2(x-2)^{3/2}}\) | 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] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | 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 \(x=11\), \(u=9\) | ||
| \(\Rightarrow \int_3^{11}\frac{x}{\sqrt{x-2}}\,dx = \int_1^9 \frac{u+2}{u^{1/2}}\,du\) | B1 | \(\int\frac{u+2}{u^{1/2}}(du)\) |
| \(= \int_1^9 (u^{1/2} + 2u^{-1/2})\,du\) | M1 | Splitting their fraction (correctly) and \(u/u^{1/2} = u^{1/2}\) (or \(\sqrt{u}\)). Or integration by parts: \(2u^{1/2}(u+2)-\int 2u^{1/2}\,du\) (must be fully correct – condone missing bracket). By parts: \([2u^{1/2}(u+2) - 4u^{3/2}/3]\) |
| \(= \left[\frac{2}{3}u^{3/2} + 4u^{1/2}\right]_1^9\) | A1 | \(\left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]\) (o.e.) |
| \(= (18+12)-(2/3+4)\) | M1 | Substituting correct limits. \(F(9)-F(1)\) (\(u\)) or \(F(11)-F(3)\) (\(x\)) |
| \(= 25\frac{1}{3}\) | 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 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=3$, $y = 3/\sqrt{(3-2)} = 3$ | M1 | Substituting $x=3$ (both $x$'s). Or $x = x/\sqrt{(x-2)}$ M1 $\Rightarrow x=3$ A1 (by solving or verifying) |
| So P is $(3,3)$ which lies on $y=x$ | A1 | $y=3$ and completion ('$3=3$' is enough) |
| **[2]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{\sqrt{x-2}\cdot1 - x\cdot\frac{1}{2}(x-2)^{-1/2}}{x-2}$ | M1 | Quotient or product rule. 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 |
| | 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 $\sqrt{(x-2)}$ o.e. e.g. taking out factor of $(x-2)^{-3/2}$. Allow ft on correct equivalent algebra from their incorrect expression |
| $= \frac{x-4}{2(x-2)^{3/2}}$ | 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]** | | |
### Part (iii)
| Answer | 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 $x=11$, $u=9$ | | |
| $\Rightarrow \int_3^{11}\frac{x}{\sqrt{x-2}}\,dx = \int_1^9 \frac{u+2}{u^{1/2}}\,du$ | B1 | $\int\frac{u+2}{u^{1/2}}(du)$ |
| $= \int_1^9 (u^{1/2} + 2u^{-1/2})\,du$ | M1 | Splitting their fraction (correctly) and $u/u^{1/2} = u^{1/2}$ (or $\sqrt{u}$). Or integration by parts: $2u^{1/2}(u+2)-\int 2u^{1/2}\,du$ (must be fully correct – condone missing bracket). By parts: $[2u^{1/2}(u+2) - 4u^{3/2}/3]$ |
| $= \left[\frac{2}{3}u^{3/2} + 4u^{1/2}\right]_1^9$ | A1 | $\left[\frac{2}{3}u^{3/2}+4u^{1/2}\right]$ (o.e.) |
| $= (18+12)-(2/3+4)$ | M1 | Substituting correct limits. $F(9)-F(1)$ ($u$) or $F(11)-F(3)$ ($x$) |
| $= 25\frac{1}{3}$ | 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]** | | |5 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]{93ee09be-f014-4dd7-a8da-8646837b17a5-2_606_732_867_710}
\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 Q5 [18]}}