| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | January |
| Marks | 10 |
| 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 logarithmic integration with substitution u=3x-1, requiring only basic technique and careful arithmetic with the limits. Part (ii) involves algebraic division to find constants A, B, C, then integrating term-by-term—standard P3 material but requires more steps. Both parts are routine textbook exercises with no novel problem-solving, making this slightly easier than average for A-level. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int \frac{2}{3x-1}dx = \frac{2}{3}\ln(3x-1)\) | M1 A1 | |
| \(\int_3^{42} \frac{2}{3x-1}dx = \frac{2}{3}\ln(125) - \frac{2}{3}\ln(8) = \ln\frac{25}{4}\) | dM1 A1 | (4) |
| (ii) \(\frac{2x^3 - 7x^2 + 8x + 1}{(x-1)^2} = 2x + B + \frac{C}{(x-1)^2}\) | B1 | |
| Full method to find values of A, B and C | M1 | |
| \(\frac{2x^3 - 7x^2 + 8x + 1}{(x-1)^2} = 2x - 3 + \frac{4}{(x-1)^2}\) | A1 | |
| \(\int h(x)dx = \int Ax + B + \frac{C}{(x-1)^2}dx = \frac{1}{2}Ax^2 + Bx - \frac{C}{(x-1)}\) | M1 A1 ft | |
| \(= x^2 - 3x - \frac{4}{(x-1)}\) \((+c)\) | A1 | (6) |
| Alt I (first 3 marks): \(2x^3 - 7x^2 + 8x + 1 = Ax(x-1)^2 + B(x-1)^2 + C\) | B1 | |
| Any of \(A = 2, B = -3, C = 4\) | M1 | |
| Either substitution or equating coefficients to get two values; All values correct \(A = 2, B = -3, C = 4\) | A1 | (3) |
| Alt II (first 3 marks): \(x^2 - 2x + 1)\overline{2x^3 - 7x^2 + 8x + 1}\) | B1 | |
| Likely to be \(2x\); For attempt at division; Correct quotient and remainder | M1 A1 | (3) |
**(i)** $\int \frac{2}{3x-1}dx = \frac{2}{3}\ln(3x-1)$ | M1 A1 |
$\int_3^{42} \frac{2}{3x-1}dx = \frac{2}{3}\ln(125) - \frac{2}{3}\ln(8) = \ln\frac{25}{4}$ | dM1 A1 | (4)
**(ii)** $\frac{2x^3 - 7x^2 + 8x + 1}{(x-1)^2} = 2x + B + \frac{C}{(x-1)^2}$ | B1 |
Full method to find values of A, B and C | M1 |
$\frac{2x^3 - 7x^2 + 8x + 1}{(x-1)^2} = 2x - 3 + \frac{4}{(x-1)^2}$ | A1 |
$\int h(x)dx = \int Ax + B + \frac{C}{(x-1)^2}dx = \frac{1}{2}Ax^2 + Bx - \frac{C}{(x-1)}$ | M1 A1 ft |
$= x^2 - 3x - \frac{4}{(x-1)}$ $(+c)$ | A1 | (6)
**Alt I (first 3 marks):** $2x^3 - 7x^2 + 8x + 1 = Ax(x-1)^2 + B(x-1)^2 + C$ | B1 |
Any of $A = 2, B = -3, C = 4$ | M1 |
Either substitution or equating coefficients to get two values; All values correct $A = 2, B = -3, C = 4$ | A1 | (3)
**Alt II (first 3 marks):** $x^2 - 2x + 1)\overline{2x^3 - 7x^2 + 8x + 1}$ | B1 |
Likely to be $2x$; For attempt at division; Correct quotient and remainder | M1 A1 | (3)
**Total: 10 marks**
---8. (i) Find, using algebraic integration, the exact value of
$$\int _ { 3 } ^ { 42 } \frac { 2 } { 3 x - 1 } \mathrm {~d} x$$
giving your answer in simplest form.\\
(ii)
$$\mathrm { h } ( x ) = \frac { 2 x ^ { 3 } - 7 x ^ { 2 } + 8 x + 1 } { ( x - 1 ) ^ { 2 } } \quad x > 1$$
Given $\mathrm { h } ( x ) = A x + B + \frac { C } { ( x - 1 ) ^ { 2 } }$ where $A , B$ and $C$ are constants to be found, find
$$\int \mathrm { h } ( x ) \mathrm { d } x$$
\includegraphics[max width=\textwidth, alt={}, center]{1c700103-ecab-4a08-b411-3f445ed88885-26_2258_47_312_1985}
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\hfill \mbox{\textit{Edexcel P3 2020 Q8 [10]}}