| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show integral transforms via substitution then evaluate (algebraic/exponential) |
| Difficulty | Standard +0.3 This is a standard C4 integration by substitution question with straightforward algebraic manipulation. Part (i) requires routine verification of a given substitution, and part (ii) involves integrating a simple rational function and evaluating definite integral limits—all well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to connect \(du\) and \(dx\) e.g. \(\frac{du}{dx} = e^x\) | M1 | But not \(du = dx\) |
| Use of \(e^{2x} = (e^x)^2\) or \((u-1)^2\) s.o.i. | A1 | |
| Simplification to \(\int\frac{u-1}{u}(du)\) WWW | A1 | 3 AG |
| Answer | Marks | Guidance |
|---|---|---|
| Change \(\frac{u-1}{u}\) to \(1 - \frac{1}{u}\) or use parts | M1 | If parts, may be twice if \(\int\ln x\,dx\) is involved |
| \(\int\frac{1}{u}du = \ln u\) | A1 | Seen anywhere in this part |
| Either attempt to change limits or resubstitute | M1 (indep) | Expect new limits e+1 & 2 |
| Show as \(e + 1 - \ln(e+1) - \{2\) or \((1+1)\} + \ln 2\) | A1 | |
| WWW show final result as \(e - 1 - \ln\left(\frac{e+1}{2}\right)\) | A1 | 5 AG |
### (i)
Attempt to connect $du$ and $dx$ e.g. $\frac{du}{dx} = e^x$ | M1 | But not $du = dx$
Use of $e^{2x} = (e^x)^2$ or $(u-1)^2$ s.o.i. | A1 |
Simplification to $\int\frac{u-1}{u}(du)$ WWW | A1 | 3 AG
---
### (ii)
Change $\frac{u-1}{u}$ to $1 - \frac{1}{u}$ or use parts | M1 | If parts, may be twice if $\int\ln x\,dx$ is involved
$\int\frac{1}{u}du = \ln u$ | A1 | Seen anywhere in this part
Either attempt to change limits or resubstitute | M1 (indep) | Expect new limits e+1 & 2
Show as $e + 1 - \ln(e+1) - \{2$ or $(1+1)\} + \ln 2$ | A1 |
WWW show final result as $e - 1 - \ln\left(\frac{e+1}{2}\right)$ | A1 | 5 AG
---\begin{enumerate}[label=(\roman*)]
\item Show that the substitution $u = e^x + 1$ transforms $\int \frac{e^{2x}}{e^x + 1} dx$ to $\int \frac{u - 1}{u} du$. [3]
\item Hence show that $\int_0^1 \frac{e^{2x}}{e^x + 1} dx = e - 1 - \ln\left(\frac{e + 1}{2}\right)$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2006 Q6 [8]}}