| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper algebraic form then partial fractions |
| Difficulty | Standard +0.3 This is a standard Further Maths partial fractions question with improper form requiring polynomial division first, followed by straightforward integration. The algebraic manipulation is routine, and the definite integral evaluation is mechanical with no conceptual challenges—slightly easier than average A-level difficulty. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + 8x - 3 = (Ax+B)(x+2) + C\) giving \(A=\ldots, B=\ldots, C=\ldots\) or polynomial long division giving quotient \(x+6\) remainder \(-15\) | M1 | Multiplies by \((x+2)\) and attempts \(A\), \(B\), \(C\); or divides to get linear quotient and constant remainder |
| Two of \(A=1, B=6, C=-15\) | A1 | |
| All three of \(A=1, B=6, C=-15\) | A1 | Implied by stating \(\frac{x^2+8x-3}{x+2} = x+6 - \frac{15}{x+2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{x^2+8x-3}{x+2}\,dx = \int x + 6 - \frac{15}{x+2}\,dx = \ldots - 15\ln(x+2)\) | M1 | Integrates expression of form \(\frac{C}{x+2}\) to obtain \(k\ln(x+2)\); condone omission of brackets |
| \(= \frac{1}{2}x^2 + 6x - 15\ln(x+2) \quad (+c)\) | A1ft | Correct integration ft on their \(Ax+B+\frac{C}{x+2}\); brackets must be present around "\(x+2\)" unless implied |
| \(\left[\frac{1}{2}x^2 + 6x - 15\ln(x+2)\right]_0^6 = (18+36-15\ln 8)-(0+0-15\ln 2)\) \(= 18+36-(15-45)\ln 2\) | M1 | Substitutes both limits into expression containing \(x\) or \(x^2\) and ln term; subtracts; combines log terms correctly |
| \(= 54 - 30\ln 2\) | A1 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + 8x - 3 = (Ax+B)(x+2) + C$ giving $A=\ldots, B=\ldots, C=\ldots$ **or** polynomial long division giving quotient $x+6$ remainder $-15$ | M1 | Multiplies by $(x+2)$ and attempts $A$, $B$, $C$; or divides to get linear quotient and constant remainder |
| Two of $A=1, B=6, C=-15$ | A1 | |
| All three of $A=1, B=6, C=-15$ | A1 | Implied by stating $\frac{x^2+8x-3}{x+2} = x+6 - \frac{15}{x+2}$ |
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# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{x^2+8x-3}{x+2}\,dx = \int x + 6 - \frac{15}{x+2}\,dx = \ldots - 15\ln(x+2)$ | M1 | Integrates expression of form $\frac{C}{x+2}$ to obtain $k\ln(x+2)$; condone omission of brackets |
| $= \frac{1}{2}x^2 + 6x - 15\ln(x+2) \quad (+c)$ | A1ft | Correct integration ft on their $Ax+B+\frac{C}{x+2}$; brackets must be present around "$x+2$" unless implied |
| $\left[\frac{1}{2}x^2 + 6x - 15\ln(x+2)\right]_0^6 = (18+36-15\ln 8)-(0+0-15\ln 2)$ $= 18+36-(15-45)\ln 2$ | M1 | Substitutes both limits into expression containing $x$ or $x^2$ and ln term; subtracts; combines log terms correctly |
| $= 54 - 30\ln 2$ | A1 | |
---\begin{enumerate}
\item (a) Given that
\end{enumerate}
$$\frac { x ^ { 2 } + 8 x - 3 } { x + 2 } \equiv A x + B + \frac { C } { x + 2 } \quad x \in \mathbb { R } \quad x \neq - 2$$
find the values of the constants $A , B$ and $C$\\
(b) Hence, using algebraic integration, find the exact value of
$$\int _ { 0 } ^ { 6 } \frac { x ^ { 2 } + 8 x - 3 } { x + 2 } d x$$
giving your answer in the form $a + b \ln 2$ where $a$ and $b$ are integers to be found.
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q6 [7]}}