| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with complex substitution requiring algebraic rearrangement |
| Difficulty | Standard +0.3 Part (i) is a standard substitution integral with u = √(2x-1) requiring routine algebraic manipulation and definite integral evaluation. Part (ii) is a straightforward partial fractions decomposition followed by logarithmic integration. Both are textbook exercises testing core A-level techniques without requiring problem-solving insight, making this slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States suitable substitution e.g. \(u=\sqrt{2x-1}\) or \(u=2x-1\) | B1 | Must be one that works |
| Changes integral in \(x\) to integral in \(u\) (all aspects including \(dx\)) | M1 | For \(u=\sqrt{2x-1}\): expect \(\int Au^2+B\,du\) |
| \(= \frac{1}{2}u^3 + \frac{3}{2}u\) or \(= \frac{1}{2}u^{3/2}+\frac{3}{2}u^{1/2}\) | dM1 A1 | |
| Applies correct limits in \(u\) | M1 | |
| \(= 16\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{6x^2-16}{(x+1)(2x-3)} = 3 + \ldots\) | B1 | |
| \(\frac{6x^2-16}{(x+1)(2x-3)} = 3 + \frac{2}{x+1} - \frac{1}{2x-3}\) | M1 A1 | |
| \(= 3x + 2\ln | x+1 | - \frac{1}{2}\ln |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States suitable substitution e.g. $u=\sqrt{2x-1}$ or $u=2x-1$ | B1 | Must be one that works |
| Changes integral in $x$ to integral in $u$ (all aspects including $dx$) | M1 | For $u=\sqrt{2x-1}$: expect $\int Au^2+B\,du$ |
| $= \frac{1}{2}u^3 + \frac{3}{2}u$ or $= \frac{1}{2}u^{3/2}+\frac{3}{2}u^{1/2}$ | dM1 A1 | |
| Applies correct limits in $u$ | M1 | |
| $= 16$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{6x^2-16}{(x+1)(2x-3)} = 3 + \ldots$ | B1 | |
| $\frac{6x^2-16}{(x+1)(2x-3)} = 3 + \frac{2}{x+1} - \frac{1}{2x-3}$ | M1 A1 | |
| $= 3x + 2\ln|x+1| - \frac{1}{2}\ln|2x-3| + c$ | M1 A1ft A1 | |7. (i) Using a suitable substitution, find, using calculus, the value of
$$\int _ { 1 } ^ { 5 } \frac { 3 x } { \sqrt { 2 x - 1 } } \mathrm {~d} x$$
(Solutions relying entirely on calculator technology are not acceptable.)\\
(ii) Find
$$\int \frac { 6 x ^ { 2 } - 16 } { ( x + 1 ) ( 2 x - 3 ) } d x$$
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\hfill \mbox{\textit{Edexcel P4 2020 Q7 [12]}}