| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Indefinite integral with linear substitution |
| Difficulty | Moderate -0.3 Part (i) is a straightforward substitution u=2x-1 leading to a standard logarithm result. Part (ii) requires recognizing that sec(x/3)tan(x/3) is the derivative of sec(x/3) and sin(2x) integrates directly to -cos(2x)/2. Both parts are routine applications of standard techniques with no novel insight required, making this slightly easier than average for P3 level. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| VIIIV SIHI NI JIIYM IONOO | VIUV SIHI NI JIIAM ION OO | VI4V SIHI NI JIIIM I ON OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{(2x-1)}dx = \frac{1}{2}\ln(2x-1)\) | M1 A1 | \(k\ln(2x-1)\) where \(k\) is a constant |
| \(\int_5^{13} \frac{1}{(2x-1)}dx = \frac{1}{2}\ln 25 - \frac{1}{2}\ln 9 = \frac{1}{2}\ln\left(\frac{25}{9}\right)\) | dM1 | Substituting limits, subtracting, using at least one log law |
| \(= \ln\left(\frac{5}{3}\right)\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrates to give \(\alpha\cos 2x + \beta\sec\frac{1}{3}x \{+c\}\) where \(\alpha \neq 0, \beta \neq 0\) | M1 | |
| \(\left[-\frac{1}{2}\cos 2x + 3\sec\frac{1}{3}x\{+c\}\right]\) | ||
| \(\left(-\frac{1}{2}\cos\left(2\times\frac{\pi}{2}\right) + 3\sec\left(\frac{1}{3}\times\frac{\pi}{2}\right)\right) - \left(-\frac{1}{2}\cos(0) + 3\sec(0)\right)\) | dM1 | Substitutes limits 0 and \(\frac{\pi}{2}\), subtracts correct way |
| \(= 2\sqrt{3} - 2\) | A1 | cao |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{(2x-1)}dx = \frac{1}{2}\ln(2x-1)$ | M1 A1 | $k\ln(2x-1)$ where $k$ is a constant |
| $\int_5^{13} \frac{1}{(2x-1)}dx = \frac{1}{2}\ln 25 - \frac{1}{2}\ln 9 = \frac{1}{2}\ln\left(\frac{25}{9}\right)$ | dM1 | Substituting limits, subtracting, using at least one log law |
| $= \ln\left(\frac{5}{3}\right)$ | A1 | cao |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrates to give $\alpha\cos 2x + \beta\sec\frac{1}{3}x \{+c\}$ where $\alpha \neq 0, \beta \neq 0$ | M1 | |
| $\left[-\frac{1}{2}\cos 2x + 3\sec\frac{1}{3}x\{+c\}\right]$ | | |
| $\left(-\frac{1}{2}\cos\left(2\times\frac{\pi}{2}\right) + 3\sec\left(\frac{1}{3}\times\frac{\pi}{2}\right)\right) - \left(-\frac{1}{2}\cos(0) + 3\sec(0)\right)$ | dM1 | Substitutes limits 0 and $\frac{\pi}{2}$, subtracts correct way |
| $= 2\sqrt{3} - 2$ | A1 | cao |
---4. (i) Find
$$\int _ { 5 } ^ { 13 } \frac { 1 } { ( 2 x - 1 ) } \mathrm { d } x$$
writing your answer in its simplest form.\\
(ii) Use integration to find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin 2 x + \sec \frac { 1 } { 3 } x \tan \frac { 1 } { 3 } x \mathrm {~d} x$$
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\hfill \mbox{\textit{Edexcel P3 2018 Q4 [7]}}