| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with partial fractions (infinite limit) |
| Difficulty | Standard +0.8 This question requires partial fractions decomposition, integration of logarithmic forms, evaluating limits at infinity, and manipulating logarithms to reach a specific form. While the techniques are standard for Further Maths Core Pure, the improper integral adds complexity beyond routine A-level integration, and the final simplification to exact form requires careful algebraic manipulation. It's moderately challenging but within expected FM scope. |
| Spec | 4.08c Improper integrals: infinite limits or discontinuous integrands4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The interval being integrated over is unbounded / the upper limit is infinity / a limit is required to evaluate it | B1 | Must refer to interval being unbounded or upper limit being infinity. Do not award for erroneous statements e.g. referring to \(x=0\). Do not accept "because one of the limits is undefined" unless they state they mean \(\infty\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{x(2x+5)} = \frac{A}{x} + \frac{B}{2x+5} \Rightarrow A=\ldots, B=\ldots\) | M1 | Selects correct form for partial fractions and finds values for \(A\) and \(B\) |
| \(\frac{1}{x(2x+5)} = \frac{1}{5x} - \frac{2}{5(2x+5)}\) | A1 | Correct constants |
| \(\int \frac{1}{5x} - \frac{2}{5(2x+5)}\,dx = \frac{1}{5}\ln x - \frac{1}{5}\ln(2x+5)\) | A1ft | \(\int \frac{p}{x} + \frac{q}{2x+5}\,dx = p\ln x + \frac{q}{2}\ln(2x+5)\). Note \(\frac{1}{5}\ln 5x - \frac{1}{5}\ln(10x+25)\) is correct |
| \(\frac{1}{5}\ln x - \frac{1}{5}\ln(2x+5) = \frac{1}{5}\ln\frac{x}{(2x+5)}\) | M1 | Combines logs correctly. May see \(-\frac{1}{5}\ln\!\left(\frac{2x+5}{x}\right) = -\frac{1}{5}\ln\!\left(2+\frac{5}{x}\right)\) |
| \(\lim_{x\to\infty}\left\{\frac{1}{5}\ln\frac{x}{2x+5}\right\} = \frac{1}{5}\ln\frac{1}{2}\) | B1 | Correct upper limit by recognising dominant terms. Simply replacing \(x\) with \(\infty\) scores B0 |
| \(\Rightarrow \int_1^{\infty} \frac{1}{x(2x+5)}\,dx = \frac{1}{5}\ln\frac{1}{2} - \frac{1}{5}\ln\frac{1}{7} = \frac{1}{5}\ln\frac{7}{2}\) | A1 | Deduces correct value in correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{x(2x+5)} = \frac{1}{2\!\left(x^2+\frac{5}{2}x\right)} = \frac{1}{2}\times\frac{1}{\left(x+\frac{5}{4}\right)^2 - \frac{25}{16}}\) | M1, A1 | Expands denominator and completes the square; correct expression |
| \(\int \frac{1}{x(2x+5)}\,dx = \frac{1}{2}\times\frac{2}{5}\ln\left\lvert\frac{x+\frac{5}{4}-\frac{5}{4}}{x+\frac{5}{4}+\frac{5}{4}}\right\rvert = \frac{1}{5}\ln\left\lvert\frac{2x}{2x+5}\right\rvert\) | M1, A1ft | For \(\frac{1}{(x+p)^2-a^2} \to k\ln\left\lvert\frac{x+p-a}{x+p+a}\right\rvert\); \(\frac{1}{2}\frac{1}{(x+a)^2-a^2}\to\frac{1}{2a}\ln\left\lvert\frac{x}{x+2a}\right\rvert\) with their \(a\) |
| \(\lim_{x\to\infty}\left\{\frac{1}{5}\ln\frac{2x}{2x+5}\right\} = \frac{1}{5}\ln\frac{2}{2} = 0\) | B1 | Correct upper limit. Note in this method the upper limit evaluates to zero |
| \(\Rightarrow \int_1^{\infty}\frac{1}{x(2x+5)}\,dx = 0 - \frac{1}{5}\ln\frac{1}{7} = \frac{1}{5}\ln\frac{7}{2}\) | A1 | Deduces correct value. Accept \(-\frac{1}{5}\ln\frac{2}{7}\) |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| The interval being integrated over is unbounded / the upper limit is infinity / a limit is required to evaluate it | B1 | Must refer to interval being unbounded or upper limit being infinity. Do not award for erroneous statements e.g. referring to $x=0$. Do not accept "because one of the limits is undefined" unless they state they mean $\infty$ |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{x(2x+5)} = \frac{A}{x} + \frac{B}{2x+5} \Rightarrow A=\ldots, B=\ldots$ | M1 | Selects correct form for partial fractions and finds values for $A$ and $B$ |
| $\frac{1}{x(2x+5)} = \frac{1}{5x} - \frac{2}{5(2x+5)}$ | A1 | Correct constants |
| $\int \frac{1}{5x} - \frac{2}{5(2x+5)}\,dx = \frac{1}{5}\ln x - \frac{1}{5}\ln(2x+5)$ | A1ft | $\int \frac{p}{x} + \frac{q}{2x+5}\,dx = p\ln x + \frac{q}{2}\ln(2x+5)$. Note $\frac{1}{5}\ln 5x - \frac{1}{5}\ln(10x+25)$ is correct |
| $\frac{1}{5}\ln x - \frac{1}{5}\ln(2x+5) = \frac{1}{5}\ln\frac{x}{(2x+5)}$ | M1 | Combines logs correctly. May see $-\frac{1}{5}\ln\!\left(\frac{2x+5}{x}\right) = -\frac{1}{5}\ln\!\left(2+\frac{5}{x}\right)$ |
| $\lim_{x\to\infty}\left\{\frac{1}{5}\ln\frac{x}{2x+5}\right\} = \frac{1}{5}\ln\frac{1}{2}$ | B1 | Correct upper limit by recognising dominant terms. Simply replacing $x$ with $\infty$ scores B0 |
| $\Rightarrow \int_1^{\infty} \frac{1}{x(2x+5)}\,dx = \frac{1}{5}\ln\frac{1}{2} - \frac{1}{5}\ln\frac{1}{7} = \frac{1}{5}\ln\frac{7}{2}$ | A1 | Deduces correct value in correct form |
### Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{x(2x+5)} = \frac{1}{2\!\left(x^2+\frac{5}{2}x\right)} = \frac{1}{2}\times\frac{1}{\left(x+\frac{5}{4}\right)^2 - \frac{25}{16}}$ | M1, A1 | Expands denominator and completes the square; correct expression |
| $\int \frac{1}{x(2x+5)}\,dx = \frac{1}{2}\times\frac{2}{5}\ln\left\lvert\frac{x+\frac{5}{4}-\frac{5}{4}}{x+\frac{5}{4}+\frac{5}{4}}\right\rvert = \frac{1}{5}\ln\left\lvert\frac{2x}{2x+5}\right\rvert$ | M1, A1ft | For $\frac{1}{(x+p)^2-a^2} \to k\ln\left\lvert\frac{x+p-a}{x+p+a}\right\rvert$; $\frac{1}{2}\frac{1}{(x+a)^2-a^2}\to\frac{1}{2a}\ln\left\lvert\frac{x}{x+2a}\right\rvert$ with their $a$ |
| $\lim_{x\to\infty}\left\{\frac{1}{5}\ln\frac{2x}{2x+5}\right\} = \frac{1}{5}\ln\frac{2}{2} = 0$ | B1 | Correct upper limit. Note in this method the upper limit evaluates to zero |
| $\Rightarrow \int_1^{\infty}\frac{1}{x(2x+5)}\,dx = 0 - \frac{1}{5}\ln\frac{1}{7} = \frac{1}{5}\ln\frac{7}{2}$ | A1 | Deduces correct value. Accept $-\frac{1}{5}\ln\frac{2}{7}$ |
---\begin{enumerate}
\item (a) Explain why $\int _ { 1 } ^ { \infty } \frac { 1 } { x ( 2 x + 5 ) } d x$ is an improper integral.\\
(b) Prove that
\end{enumerate}
$$\int _ { 1 } ^ { \infty } \frac { 1 } { x ( 2 x + 5 ) } d x = a \ln b$$
where $a$ and $b$ are rational numbers to be determined.
\hfill \mbox{\textit{Edexcel CP1 2020 Q2 [7]}}