OCR MEI C3 — Question 2 18 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeMulti-part questions combining substitution with curve/area analysis
DifficultyStandard +0.8 This is a multi-part question requiring algebraic manipulation with substitution, polynomial division, integration by parts, and connecting results across parts. While each technique is standard C3 material, the need to perform polynomial division to obtain the specific form, then later combine integration by parts with the earlier result, requires more sophistication than typical textbook exercises. The multi-step reasoning and synthesis across parts elevates this above average difficulty.
Spec1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08d Evaluate definite integrals: between limits1.08h Integration by substitution1.08i Integration by parts
2
  1. 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]
    \includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_706} \captionsetup{labelformat=empty} \caption{Fig. 8}
    \end{figure}
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Verify that the origin is a stationary point of the curve.
  3. 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 \(x = 1\).
Question 2:
Part (i): \(\int_0^1 \frac{x^3}{1+x}\,dx\), let \(u = 1+x\), \(du = dx\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(a=1, b=2\) and \((u-1)^3/u\)B1, B1 Seen anywhere, e.g. in new limits
Expanding \((u^3 - 3u^2 + 3u - 1)/u\) correctlyM1 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 \(du=1\)
\(\left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]_1^2\)B1 Correct integration
Substituting correct limits dep integratedM1 upper − lower; may be implied from 0.140…
\(= \frac{5}{6} - \ln 2\)A1cao [7] Must be exact − must be 5/6; must have evaluated \(\ln 1 = 0\)
Part (ii): \(y = x^2\ln(1+x)\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Product ruleM1
\(\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}\)B1 or \(\frac{d}{dx}(\ln u) = 1/u\) where \(u=1+x\); \(\ln 1 + x\) is A0
\(\frac{dy}{dx} = \frac{x^2}{1+x} + 2x\ln(1+x)\)A1cao [5]
Substituting \(x=0\) into correct deriv; \(dy/dx = 0\)M1, A1cao when \(x=0\), \(dy/dx = 0\) with no evidence of substituting M1A0; condone missing bracket in \(\ln(1+x)\)
Origin is a stationary point[5]
Part (iii): \(A = \int_0^1 x^2\ln(1+x)\,dx\)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct integral and limitsB1 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)
\(A = \left[\frac{1}{3}x^3\ln(1+x)\right]_0^1 - \int_0^1 \frac{1}{3}\cdot\frac{x^3}{1+x}\,dx\)A1 Condone missing brackets
\(= \frac{1}{3}\ln 2 - \ldots\)B1 \(= \frac{1}{3}\ln 2 - \ldots\)
\(= \frac{1}{3}\ln 2 - \frac{5}{18} + \frac{1}{3}\ln 2\)B1ft \(\ldots - 1/3\) (result from part (i)); condone missing bracket, can re-work from scratch
\(= \frac{2}{3}\ln 2 - \frac{5}{18}\)A1 [6] cao; oe e.g. \(\frac{12\ln 2 - 5}{18}\), \(\frac{1}{3}\ln 4 - \frac{5}{18}\); must have evaluated \(\ln 1 = 0\); must combine the two ln terms 
## Question 2:

### Part (i): $\int_0^1 \frac{x^3}{1+x}\,dx$, let $u = 1+x$, $du = dx$

| Answer/Working | Marks | Guidance |
|---|---|---|
| $a=1, b=2$ and $(u-1)^3/u$ | B1, B1 | Seen anywhere, e.g. in new limits |
| Expanding $(u^3 - 3u^2 + 3u - 1)/u$ correctly | 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 $du=1$ |
| $\left[\frac{1}{3}u^3 - \frac{3}{2}u^2 + 3u - \ln u\right]_1^2$ | B1 | Correct integration |
| Substituting correct limits dep integrated | M1 | upper − lower; may be implied from 0.140… |
| $= \frac{5}{6} - \ln 2$ | A1cao [7] | Must be exact − must be 5/6; must have evaluated $\ln 1 = 0$ |

### Part (ii): $y = x^2\ln(1+x)$

| Answer/Working | Marks | Guidance |
|---|---|---|
| Product rule | M1 | |
| $\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}$ | B1 | or $\frac{d}{dx}(\ln u) = 1/u$ where $u=1+x$; $\ln 1 + x$ is A0 |
| $\frac{dy}{dx} = \frac{x^2}{1+x} + 2x\ln(1+x)$ | A1cao [5] | |
| Substituting $x=0$ into **correct** deriv; $dy/dx = 0$ | M1, A1cao | when $x=0$, $dy/dx = 0$ with no evidence of substituting M1A0; condone missing bracket in $\ln(1+x)$ |
| Origin is a stationary point | [5] | |

### Part (iii): $A = \int_0^1 x^2\ln(1+x)\,dx$

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct integral and limits | B1 | 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) |
| $A = \left[\frac{1}{3}x^3\ln(1+x)\right]_0^1 - \int_0^1 \frac{1}{3}\cdot\frac{x^3}{1+x}\,dx$ | A1 | Condone missing brackets |
| $= \frac{1}{3}\ln 2 - \ldots$ | B1 | $= \frac{1}{3}\ln 2 - \ldots$ |
| $= \frac{1}{3}\ln 2 - \frac{5}{18} + \frac{1}{3}\ln 2$ | B1ft | $\ldots - 1/3$ (result from part (i)); condone missing bracket, can re-work from scratch |
| $= \frac{2}{3}\ln 2 - \frac{5}{18}$ | A1 [6] | cao; oe e.g. $\frac{12\ln 2 - 5}{18}$, $\frac{1}{3}\ln 4 - \frac{5}{18}$; must have evaluated $\ln 1 = 0$; must combine the two ln terms |

---
2 (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]{35646966-3747-4f1d-bf94-60e9e3130afe-2_829_806_944_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  Q2 [18]}}
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