| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Derivative then integrate by parts |
| Difficulty | Standard +0.3 This is a guided integration by parts question where part (i) provides the key insight by asking students to differentiate the answer first. Once they see the derivative simplifies to 5e^x sin 2x, part (ii) becomes straightforward evaluation of definite integral limits. The structure removes the problem-solving challenge of choosing u and dv, making it slightly easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate as a product, \(u\,dv + v\,du\) | M1 | or as 2 separate products |
| \(\frac{d}{dx}(\sin 2x)=2\cos 2x\) or \(\frac{d}{dx}(\cos 2x)=-2\sin 2x\) | B1 | |
| \(e^x(2\cos 2x+4\sin 2x)+e^x(\sin 2x-2\cos 2x)\) | A1 | terms may be in diff order |
| Simplify to \(5e^x\sin 2x\) | A1 4 | Accept \(10e^x\sin x\cos x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Provided result (i) is of form \(ke^x\sin 2x\), \(k\) const | ||
| \(\int e^x\sin 2x\,dx = \frac{1}{k}e^x(\sin 2x-2\cos 2x)\) | B1 | |
| \(\left[e^x(\sin 2x-2\cos 2x)\right]_0^{\frac{1}{2}\pi}=e^{\frac{1}{2}\pi}+2\) | B1 | |
| \(\frac{1}{5}\left(e^{\frac{1}{2}\pi}+2\right)\) | B1 3 | Exact form to be seen |
# Question 4:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate as a product, $u\,dv + v\,du$ | M1 | or as 2 separate products |
| $\frac{d}{dx}(\sin 2x)=2\cos 2x$ or $\frac{d}{dx}(\cos 2x)=-2\sin 2x$ | B1 | |
| $e^x(2\cos 2x+4\sin 2x)+e^x(\sin 2x-2\cos 2x)$ | A1 | terms may be in diff order |
| Simplify to $5e^x\sin 2x$ | A1 **4** | Accept $10e^x\sin x\cos x$ |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Provided result (i) is of form $ke^x\sin 2x$, $k$ const | | |
| $\int e^x\sin 2x\,dx = \frac{1}{k}e^x(\sin 2x-2\cos 2x)$ | B1 | |
| $\left[e^x(\sin 2x-2\cos 2x)\right]_0^{\frac{1}{2}\pi}=e^{\frac{1}{2}\pi}+2$ | B1 | |
| $\frac{1}{5}\left(e^{\frac{1}{2}\pi}+2\right)$ | B1 **3** | Exact form to be seen |
---4 (i) Differentiate $\mathrm { e } ^ { x } ( \sin 2 x - 2 \cos 2 x )$, simplifying your answer.\\
(ii) Hence find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { e } ^ { x } \sin 2 x \mathrm {~d} x$.
\hfill \mbox{\textit{OCR C4 2009 Q4 [7]}}