| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: product rule differentiation, second derivative test, algebraic substitution in integration, and integration by parts. Each part follows directly from the previous with clear guidance. While it requires competent execution of multiple techniques (19 marks total), there are no conceptual surprises or novel problem-solving demands—slightly above average due to length and the integration by parts application, but well within typical C3 scope. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(y = 2x\ln(1+x)\); \(\frac{dy}{dx} = \frac{2x}{1+x} + 2\ln(1+x)\); When \(x = 0\), \(dy/dx = 0 + 2\ln 1 = 0\) \(\Rightarrow\) origin is a stationary point. | M1, B1, A1, E1, [4] | product rule; \(d/dx(\ln(1+x)) = 1/(1+x)\) soi; www (i.e. from correct derivative) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\frac{d^2y}{dx^2} = \frac{(1+x).2 - 2x.1}{(1+x)^2} + \frac{2}{1+x} = \frac{2}{(1+x)^2} + \frac{4+2x}{1+x}\); When \(x = 0\), \(d^2y/dx^2 = 2 + 2 = 4 > 0\) \(\Rightarrow\) \((0,0)\) is a min point | M1, A1ft, M1, E1, [5] | Quotient or product rule on their \(2x/(1+x)\) correctly applied to their \(2x/(1+x)\); o.e.e. \(\frac{4+2x}{(1+x)^2}\) cao; substituting \(x = 0\) into their \(d^2y/dx^2\); www – dep previous A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Let \(u = 1 + x \Rightarrow du = dx\); \(\int \frac{x^2}{1+x}dx = \int \frac{(u-1)^2}{u}du = \int \frac{(u-1)^2}{u}du = \int(u - 2 + \frac{1}{u})du *\) | M1, E1 | \(\frac{(u-1)^2}{u}\); www (but condone \(du\) omitted except in final answer) |
| Answer/Working: \(\Rightarrow \int_0^1 \frac{x^2}{1+x}dx = \int_1^2(u - 2 - \frac{1}{u})du = [\frac{1}{2}u^2 - 2u + \ln u]_1^2\) | B1, B1 | changing limits (or substituting back for \(x\) and using 0 and 1); \([\frac{1}{2}u^2 - 2u + \ln u]\) |
| Answer/Working: \(= 2 - 4 + \ln 2 - (\frac{1}{2} - 2 + \ln 1) = \ln 2 - \frac{1}{2}\) | M1, A1, [6] | substituting limits (consistent with \(u\) or \(x\)) cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(A = \int_0^1 2x\ln(1+x)dx\); Parts: \(u = \ln(1+x)\), \(du/dx = 1/(1+x)\), \(dv/dx = 2x \Rightarrow v = x^2\) | M1, A1, M1 | soi; substituting their \(\ln 2 - \frac{1}{2}\) for \(\int_0^1 \frac{x^2}{1+x}dx\) |
| Answer/Working: \(= [x^2\ln(1+x)]_0^1 - \int_0^1 \frac{x^2}{1+x}dx = \ln 2 - \ln 2 + \frac{1}{2} = \frac{1}{2}\) | A1, M1, A1, [4] | cao |
## Part (i)
**Answer/Working:** $y = 2x\ln(1+x)$; $\frac{dy}{dx} = \frac{2x}{1+x} + 2\ln(1+x)$; When $x = 0$, $dy/dx = 0 + 2\ln 1 = 0$ $\Rightarrow$ origin is a stationary point. | **M1, B1, A1, E1, [4]** | product rule; $d/dx(\ln(1+x)) = 1/(1+x)$ soi; www (i.e. from correct derivative)
## Part (ii)
**Answer/Working:** $\frac{d^2y}{dx^2} = \frac{(1+x).2 - 2x.1}{(1+x)^2} + \frac{2}{1+x} = \frac{2}{(1+x)^2} + \frac{4+2x}{1+x}$; When $x = 0$, $d^2y/dx^2 = 2 + 2 = 4 > 0$ $\Rightarrow$ $(0,0)$ is a min point | **M1, A1ft, M1, E1, [5]** | Quotient or product rule on their $2x/(1+x)$ correctly applied to their $2x/(1+x)$; o.e.e. $\frac{4+2x}{(1+x)^2}$ cao; substituting $x = 0$ into their $d^2y/dx^2$; www – dep previous A1
## Part (iii)
**Answer/Working:** Let $u = 1 + x \Rightarrow du = dx$; $\int \frac{x^2}{1+x}dx = \int \frac{(u-1)^2}{u}du = \int \frac{(u-1)^2}{u}du = \int(u - 2 + \frac{1}{u})du *$ | **M1, E1** | $\frac{(u-1)^2}{u}$; www (but condone $du$ omitted except in final answer)
**Answer/Working:** $\Rightarrow \int_0^1 \frac{x^2}{1+x}dx = \int_1^2(u - 2 - \frac{1}{u})du = [\frac{1}{2}u^2 - 2u + \ln u]_1^2$ | **B1, B1** | changing limits (or substituting back for $x$ and using 0 and 1); $[\frac{1}{2}u^2 - 2u + \ln u]$
**Answer/Working:** $= 2 - 4 + \ln 2 - (\frac{1}{2} - 2 + \ln 1) = \ln 2 - \frac{1}{2}$ | **M1, A1, [6]** | substituting limits (consistent with $u$ or $x$) cao
## Part (iv)
**Answer/Working:** $A = \int_0^1 2x\ln(1+x)dx$; Parts: $u = \ln(1+x)$, $du/dx = 1/(1+x)$, $dv/dx = 2x \Rightarrow v = x^2$ | **M1, A1, M1** | soi; substituting their $\ln 2 - \frac{1}{2}$ for $\int_0^1 \frac{x^2}{1+x}dx$
**Answer/Working:** $= [x^2\ln(1+x)]_0^1 - \int_0^1 \frac{x^2}{1+x}dx = \ln 2 - \ln 2 + \frac{1}{2} = \frac{1}{2}$ | **A1, M1, A1, [4]** | cao
---A curve is defined by the equation $y = 2x \ln(1 + x)$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$ and hence verify that the origin is a stationary point of the curve. [4]
\item Find $\frac{d^2y}{dx^2}$ and use this to verify that the origin is a minimum point. [5]
\item Using the substitution $u = 1 + x$, show that $\int \frac{x^2}{1+x} \, dx = \int \left(u - 2 + \frac{1}{u}\right) du$.
Hence evaluate $\int_0^1 \frac{x^2}{1+x} \, dx$, giving your answer in an exact form. [6]
\item Using integration by parts and your answer to part (iii), evaluate $\int_0^1 2x \ln(1 + x) \, dx$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q3 [19]}}