| 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 | Challenging +1.2 This is a structured multi-part question requiring substitution with polynomial division, differentiation using the product rule, and integration by parts combined with a previous result. While it involves multiple techniques and careful algebraic manipulation across three connected parts, each individual step follows standard C3 procedures without requiring novel insight. The scaffolding provided (showing the transformed integral form, verifying a stationary point) reduces the problem-solving demand, making this moderately harder than average but well within typical C3 scope. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07q Product and quotient rules: differentiation1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 \frac{x^3}{1+x}dx\), let \(u = 1+x\), \(du = dx\) | ||
| When \(x=0\), \(u=1\); when \(x=1\), \(u=2\) | B1 | \(a=1\), \(b=2\); seen anywhere e.g. in new limits |
| \(\frac{(u-1)^3}{u}\) | B1 | |
| \(= \int_1^2 \frac{(u-1)^3}{u}du\) | ||
| \(= \int_1^2 \frac{u^3 - 3u^2 + 3u - 1}{u}du\) | M1 | Expanding correctly |
| \(= \int_1^2 \left(u^2 - 3u + 3 - \frac{1}{u}\right)du\) | A1dep | dep \(du = dx\) (o.e.) AG; e.g. \(du/dx = 1\), condone missing \(dx\)'s and \(du\)'s, allow \(du=1\) |
| \(\int_0^1 \frac{x^3}{1+x}dx = \left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]_1^2\) | B1 | \(\left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]\) |
| \(= \left(\frac{8}{3} - 6 + 6 - \ln 2\right) - \left(\frac{1}{3} - \frac{3}{2} + 3 - \ln 1\right)\) | M1 | Substituting correct limits dep integrated; upper \(-\) lower; may be implied from \(0.140\ldots\) |
| \(= \frac{5}{6} - \ln 2\) | A1cao | Must be exact; must be \(5/6\); must have evaluated \(\ln 1 = 0\) |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^2\ln(1+x)\) | M1 | Product rule |
| \(\Rightarrow \frac{dy}{dx} = x^2 \cdot \frac{1}{1+x} + 2x\ln(1+x)\) | B1 | \(d/dx(\ln(1+x)) = 1/(1+x)\); or \(d/dx(\ln u) = 1/u\) where \(u = 1+x\); \(\ln 1+x\) is A0 |
| \(= \frac{x^2}{1+x} + 2x\ln(1+x)\) | A1 | cao (oe) mark final answer |
| When \(x=0\), \(dy/dx = 0 + 0.\ln 1 = 0\) | M1 | Substituting \(x=0\) into correct deriv www; when \(x=0\), \(dy/dx=0\) with no evidence of substituting M1A0; but condone missing bracket in \(\ln(1+x)\) |
| \((\Rightarrow\) Origin is a stationary point) | A1cao | www |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \int_0^1 x^2\ln(1+x)dx\) | B1 | Correct integral and limits; condone no \(dx\); limits (and integral) can be implied by subsequent work |
| Let \(u = \ln(1+x)\), \(dv/dx = x^2\) | ||
| \(\frac{du}{dx} = \frac{1}{1+x}\), \(v = \frac{1}{3}x^3\) | M1 | Parts correct; \(u\), \(du/dx\), \(dv/dx\) and \(v\) all correct (oe) |
| \(\Rightarrow A = \left[\frac{1}{3}x^3\ln(1+x)\right]_0^1 - \int_0^1 \frac{1}{3}\cdot\frac{x^3}{1+x}dx\) | A1 | Condone missing brackets |
| \(= \frac{1}{3}\ln 2 - \left(\frac{5}{18} - \frac{1}{3}\ln 2\right)\) | B1 | \(= \frac{1}{3}\ln 2 - \ldots\) |
| B1ft | \(\ldots - 1/3\) (result from part (i)); condone missing bracket, can re-work from scratch | |
| \(= \frac{1}{3}\ln 2 - \frac{5}{18} + \frac{1}{3}\ln 2\) | ||
| \(= \frac{2}{3}\ln 2 - \frac{5}{18}\) | A1 | cao; o.e. e.g. \(= \frac{12\ln 2 - 5}{18}\), \(\frac{1}{3}\ln 4 - \frac{5}{18}\), etc; but must have evaluated \(\ln 1 = 0\); must combine the two \(\ln\) terms |
| [6] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 \frac{x^3}{1+x}dx$, let $u = 1+x$, $du = dx$ | | |
| When $x=0$, $u=1$; when $x=1$, $u=2$ | B1 | $a=1$, $b=2$; seen anywhere e.g. in new limits |
| $\frac{(u-1)^3}{u}$ | B1 | |
| $= \int_1^2 \frac{(u-1)^3}{u}du$ | | |
| $= \int_1^2 \frac{u^3 - 3u^2 + 3u - 1}{u}du$ | M1 | Expanding correctly |
| $= \int_1^2 \left(u^2 - 3u + 3 - \frac{1}{u}\right)du$ | A1dep | dep $du = dx$ (o.e.) **AG**; e.g. $du/dx = 1$, condone missing $dx$'s and $du$'s, allow $du=1$ |
| $\int_0^1 \frac{x^3}{1+x}dx = \left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]_1^2$ | B1 | $\left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]$ |
| $= \left(\frac{8}{3} - 6 + 6 - \ln 2\right) - \left(\frac{1}{3} - \frac{3}{2} + 3 - \ln 1\right)$ | M1 | Substituting correct limits dep integrated; upper $-$ lower; may be implied from $0.140\ldots$ |
| $= \frac{5}{6} - \ln 2$ | A1cao | Must be exact; must be $5/6$; must have evaluated $\ln 1 = 0$ |
| | **[7]** | |
---
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^2\ln(1+x)$ | M1 | Product rule |
| $\Rightarrow \frac{dy}{dx} = x^2 \cdot \frac{1}{1+x} + 2x\ln(1+x)$ | B1 | $d/dx(\ln(1+x)) = 1/(1+x)$; or $d/dx(\ln u) = 1/u$ where $u = 1+x$; $\ln 1+x$ is A0 |
| $= \frac{x^2}{1+x} + 2x\ln(1+x)$ | A1 | cao (oe) mark final answer |
| When $x=0$, $dy/dx = 0 + 0.\ln 1 = 0$ | M1 | Substituting $x=0$ into correct deriv www; when $x=0$, $dy/dx=0$ with no evidence of substituting M1A0; but condone missing bracket in $\ln(1+x)$ |
| $(\Rightarrow$ Origin is a stationary point) | A1cao | www |
| | **[5]** | |
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \int_0^1 x^2\ln(1+x)dx$ | B1 | Correct integral and limits; condone no $dx$; limits (and integral) can be implied by subsequent work |
| Let $u = \ln(1+x)$, $dv/dx = x^2$ | | |
| $\frac{du}{dx} = \frac{1}{1+x}$, $v = \frac{1}{3}x^3$ | M1 | Parts correct; $u$, $du/dx$, $dv/dx$ and $v$ all correct (oe) |
| $\Rightarrow A = \left[\frac{1}{3}x^3\ln(1+x)\right]_0^1 - \int_0^1 \frac{1}{3}\cdot\frac{x^3}{1+x}dx$ | A1 | Condone missing brackets |
| $= \frac{1}{3}\ln 2 - \left(\frac{5}{18} - \frac{1}{3}\ln 2\right)$ | B1 | $= \frac{1}{3}\ln 2 - \ldots$ |
| | B1ft | $\ldots - 1/3$ (result from part (i)); condone missing bracket, can re-work from scratch |
| $= \frac{1}{3}\ln 2 - \frac{5}{18} + \frac{1}{3}\ln 2$ | | |
| $= \frac{2}{3}\ln 2 - \frac{5}{18}$ | A1 | cao; o.e. e.g. $= \frac{12\ln 2 - 5}{18}$, $\frac{1}{3}\ln 4 - \frac{5}{18}$, etc; but must have evaluated $\ln 1 = 0$; must combine the two $\ln$ terms |
| | **[6]** | |3 (i) Use the substitution $u = 1 + x$ to show that
$$\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x = \int _ { a } ^ { b } \left( u ^ { 2 } - 3 u + 3 - \frac { 1 } { u } \right) \mathrm { d } u$$
where $a$ and $b$ are to be found.\\
Hence evaluate $\int _ { 0 } ^ { 1 } \frac { x ^ { 3 } } { 1 + x } \mathrm {~d} x$, giving your answer in exact form.
Fig. 8 shows the curve $y = x ^ { 2 } \ln ( 1 + x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-3_830_806_907_706}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
Verify that the origin is a stationary point of the curve.\\
(iii) Using integration by parts, and the result of part (i), find the exact area enclosed by the curve $y = x ^ { 2 } \ln ( 1 + x )$, the $x$-axis and the line $x = 1$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [18]}}