| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| 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 testing standard C3 techniques: finding intersection points, verifying inverse functions, integrating exponentials, and integration by parts for ln(x). While part (v) requires recognizing the symmetry property of inverse functions to find the enclosed area, each individual step follows routine procedures with no novel problem-solving required. Slightly easier than average due to heavy scaffolding. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(1 + \ln x = 0 \Rightarrow \ln x = -1\) so \(A\) is \((e^{-1}, 0)\); \(x = e^{-1}\) | M1, A1, B1 | SC1 if obtained using symmetry condone use of symmetry. Penalise \(A = e^{-1}\); \(B = e^{-1}\); or co-ords wrong way round, but condone labelling errors. |
| Answer/Working: \(B: x = 0, y = e^{0-1} = e^{-1}\) so \(B\) is \((0, e^{-1})\) | A1, B1 | |
| Answer/Working: \(f(1) = e^{1-1} = e^0 = 1\); \(g(1) = 1 + \ln 1 = 1\) | E1, E1, [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Either by inversion: \(y = e^{x-1} \Rightarrow x \Leftrightarrow y\); \(x = e^{y-1} \Rightarrow \ln x = y - 1 \Rightarrow y = 1 + \ln x = y\) | M1, E1 | taking ln or exps |
| Answer/Working: or by composing: \(f(g(x)) = e^{(1+\ln x)-1} = e^{\ln x} = x\) | M1, E1 | \(e^{1+\ln x-1}\) or \(1 + \ln(e^{x-1})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\int_0^e e^{x-1}dx = [e^{x-1}]_0^e = e^0 - e^{-1} = 1 - e^{-1}\) | M1, A1cao, [3] | \([e^{x-1}]\) o.e or \(u = x - 1 \Rightarrow [e^u]\); substituting correct limits for \(x\) or \(u\) o.e. not \(e^0\), must be exact. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: \(\int \ln xdx = \int \ln x \frac{d}{dx}(x)dx = x\ln x - \int x \cdot \frac{1}{x}dx = x\ln x - x + c\) | M1, A1 | parts: \(u = \ln x\), \(du/dx = 1/x\), \(v = x\), \(dv/dx = 1\) |
| Answer/Working: \(\Rightarrow \int_c^1 g(x)dx = \int_c^1 (1+\ln x)dx = [x + x\ln x - x]_c^1 = [x(ln x)]_{c}^1\) | B1ft | condone no 'c'; ft their 'x ln x - x' (provided 'algebraic') |
| Answer/Working: \(= \ln 1 - e^{-1}\ln(e^{-1}) = e^{-1} *\) | DM1, E1, [6] | substituting limits dep B1; www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Area \(= \int_0^1 f(x)dx - \int_c^1 g(x)dx = (1-e^{-1}) - e^{-1} = 1 - 2/e\) | M1, A1cao | Must have correct limits; 0.264 or better. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Area \(OCB =\) area under curve \(-\) triangle \(= 1 - e^{-1} - \frac{1}{2} \times 1 \times 1 = \frac{1}{2} - e^{-1}\) | M1 | OCA or OCB \(= \frac{1}{2} - e^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working: Total area \(= 2(\frac{1}{2} - e^{-1}) = 1 - 2/e\) | A1cao, [2] | 0.264 or better |
## Part (i)
**Answer/Working:** $1 + \ln x = 0 \Rightarrow \ln x = -1$ so $A$ is $(e^{-1}, 0)$; $x = e^{-1}$ | **M1, A1, B1** | SC1 if obtained using symmetry condone use of symmetry. Penalise $A = e^{-1}$; $B = e^{-1}$; or co-ords wrong way round, but condone labelling errors.
**Answer/Working:** $B: x = 0, y = e^{0-1} = e^{-1}$ so $B$ is $(0, e^{-1})$ | **A1, B1** |
**Answer/Working:** $f(1) = e^{1-1} = e^0 = 1$; $g(1) = 1 + \ln 1 = 1$ | **E1, E1, [5]** |
## Part (ii)
**Answer/Working:** Either by inversion: $y = e^{x-1} \Rightarrow x \Leftrightarrow y$; $x = e^{y-1} \Rightarrow \ln x = y - 1 \Rightarrow y = 1 + \ln x = y$ | **M1, E1** | taking ln or exps
**Answer/Working:** or by composing: $f(g(x)) = e^{(1+\ln x)-1} = e^{\ln x} = x$ | **M1, E1** | $e^{1+\ln x-1}$ or $1 + \ln(e^{x-1})$
## Part (iii)
**Answer/Working:** $\int_0^e e^{x-1}dx = [e^{x-1}]_0^e = e^0 - e^{-1} = 1 - e^{-1}$ | **M1, A1cao, [3]** | $[e^{x-1}]$ o.e or $u = x - 1 \Rightarrow [e^u]$; substituting correct limits for $x$ or $u$ o.e. not $e^0$, must be exact.
## Part (iv)
**Answer/Working:** $\int \ln xdx = \int \ln x \frac{d}{dx}(x)dx = x\ln x - \int x \cdot \frac{1}{x}dx = x\ln x - x + c$ | **M1, A1** | parts: $u = \ln x$, $du/dx = 1/x$, $v = x$, $dv/dx = 1$
**Answer/Working:** $\Rightarrow \int_c^1 g(x)dx = \int_c^1 (1+\ln x)dx = [x + x\ln x - x]_c^1 = [x(ln x)]_{c}^1$ | **B1ft** | condone no 'c'; ft their 'x ln x - x' (provided 'algebraic')
**Answer/Working:** $= \ln 1 - e^{-1}\ln(e^{-1}) = e^{-1} *$ | **DM1, E1, [6]** | substituting limits dep B1; www
## Part (v)
**Answer/Working:** Area $= \int_0^1 f(x)dx - \int_c^1 g(x)dx = (1-e^{-1}) - e^{-1} = 1 - 2/e$ | **M1, A1cao** | Must have correct limits; 0.264 or better.
**or**
**Answer/Working:** Area $OCB =$ area under curve $-$ triangle $= 1 - e^{-1} - \frac{1}{2} \times 1 \times 1 = \frac{1}{2} - e^{-1}$ | **M1** | OCA or OCB $= \frac{1}{2} - e^{-1}$
**Answer/Working:** or Area $OAC =$ triangle $-$ area under curve $= \frac{1}{2} \times 1 \times 1 - e^{-1} = \frac{1}{2} - e^{-1}$
**Answer/Working:** Total area $= 2(\frac{1}{2} - e^{-1}) = 1 - 2/e$ | **A1cao, [2]** | 0.264 or better
---Fig. 8 shows the line $y = x$ and parts of the curves $y = f(x)$ and $y = g(x)$, where
$$f(x) = e^{x-1}, \quad g(x) = 1 + \ln x.$$
The curves intersect the axes at the points A and B, as shown. The curves and the line $y = x$ meet at the point C.
\includegraphics{figure_8}
\begin{enumerate}[label=(\roman*)]
\item Find the exact coordinates of A and B. Verify that the coordinates of C are $(1, 1)$. [5]
\item Prove algebraically that $g(x)$ is the inverse of $f(x)$. [2]
\item Evaluate $\int_0^1 f(x) \, dx$, giving your answer in terms of $e$. [3]
\item Use integration by parts to find $\int \ln x \, dx$.
Hence show that $\int_{e^{-1}}^1 g(x) \, dx = \frac{1}{e}$. [6]
\item Find the area of the region enclosed by the lines OA and OB, and the arcs AC and BC. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}