| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (algebraic/exponential substitution) |
| Difficulty | Standard +0.3 This is a straightforward integration by substitution with clear guidance (the substitution is given). Students need to find du/dx = -sin x, change limits (x=0 gives u=2, x=π/2 gives u=1), integrate e^u to get [e^u], and evaluate. The algebra is clean and the question is slightly easier than average since the substitution is provided and the integration is standard. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{du}{dx} = -\sin x\) | B1 | |
| \(\int e^{\cos x+1} \sin x \, dx = -\int e^u \, du\) | M1 A1 | |
| \(= -e^u (+c)\) | A1 | ft on sign error |
| \(= -e^{(\cos x+1)}(+c)\) | ||
| \(\left[-e^{(\cos x+1)}\right]_0^{\frac{\pi}{2}} = (-e)-(-e^2) = e(e-1)\) | M1 A1* | cso |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{du}{dx} = -\sin x$ | B1 | |
| $\int e^{\cos x+1} \sin x \, dx = -\int e^u \, du$ | M1 A1 | |
| $= -e^u (+c)$ | A1 | ft on sign error |
| $= -e^{(\cos x+1)}(+c)$ | | |
| $\left[-e^{(\cos x+1)}\right]_0^{\frac{\pi}{2}} = (-e)-(-e^2) = e(e-1)$ | M1 A1* | cso |
---\begin{enumerate}
\item Using the substitution $u = \cos x + 1$, or otherwise, show that
\end{enumerate}
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { ( \cos x + 1 ) } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$
\hfill \mbox{\textit{Edexcel C34 Q3 [6]}}