AQA FP3 2009 June — Question 4 5 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypeImproper integrals with infinite upper limit (combined rational terms)
DifficultyStandard +0.8 This is a Further Maths FP3 question combining improper integrals with logarithmic integration. While the integration itself is straightforward (two standard log forms), students must correctly handle the infinite limit using a limiting process, evaluate the resulting logarithmic expression at infinity (requiring understanding that log terms vanish), and simplify to ln k form. The multi-step nature and requirement for rigorous limiting notation elevates this above routine C3/C4 work, though it's still a standard FP3 exercise without novel insight.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands
4 Evaluate the improper integral $$\int _ { 1 } ^ { \infty } \left( \frac { 1 } { x } - \frac { 4 } { 4 x + 1 } \right) \mathrm { d } x$$ showing the limiting process used and giving your answer in the form \(\ln k\), where \(k\) is a constant to be found.
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int\left(\frac{1}{x} - \frac{4}{4x+1}\right)dx = \ln x - \ln(4x+1)\{+c\}\)B1 OE
\(I = \lim_{a\to\infty}\int_1^a \left(\frac{1}{x} - \frac{4}{4x+1}\right)dx\)M1 \(\infty\) replaced by \(a\) (OE) and \(\lim_{a\to\infty}\)
\(= \lim_{a\to\infty}\left[\ln x - \ln(4x+1)\right]_1^a\)
\(= \lim_{a\to\infty}\left[\ln\left(\frac{a}{4a+1}\right) - \ln\frac{1}{5}\right]\)m1 \(\ln a - \ln(4a+1) = \ln\left(\frac{a}{4a+1}\right)\) and previous M1 scored
\(= \lim_{a\to\infty}\left[\ln\left(\frac{1}{4+\frac{1}{a}}\right) - \ln\frac{1}{5}\right]\)m1 \(\ln\left(\frac{a}{4a+1}\right) = \ln\left(\frac{1}{4+\frac{1}{a}}\right)\) and previous M1m1 scored
\(= \ln\frac{1}{4} - \ln\frac{1}{5} = \ln\frac{5}{4}\)A1 Total: 5 marks. CSO 
# Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\left(\frac{1}{x} - \frac{4}{4x+1}\right)dx = \ln x - \ln(4x+1)\{+c\}$ | B1 | OE |
| $I = \lim_{a\to\infty}\int_1^a \left(\frac{1}{x} - \frac{4}{4x+1}\right)dx$ | M1 | $\infty$ replaced by $a$ (OE) and $\lim_{a\to\infty}$ |
| $= \lim_{a\to\infty}\left[\ln x - \ln(4x+1)\right]_1^a$ | | |
| $= \lim_{a\to\infty}\left[\ln\left(\frac{a}{4a+1}\right) - \ln\frac{1}{5}\right]$ | m1 | $\ln a - \ln(4a+1) = \ln\left(\frac{a}{4a+1}\right)$ **and** previous M1 scored |
| $= \lim_{a\to\infty}\left[\ln\left(\frac{1}{4+\frac{1}{a}}\right) - \ln\frac{1}{5}\right]$ | m1 | $\ln\left(\frac{a}{4a+1}\right) = \ln\left(\frac{1}{4+\frac{1}{a}}\right)$ **and** previous M1m1 scored |
| $= \ln\frac{1}{4} - \ln\frac{1}{5} = \ln\frac{5}{4}$ | A1 | Total: 5 marks. CSO |

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4 Evaluate the improper integral

$$\int _ { 1 } ^ { \infty } \left( \frac { 1 } { x } - \frac { 4 } { 4 x + 1 } \right) \mathrm { d } x$$

showing the limiting process used and giving your answer in the form $\ln k$, where $k$ is a constant to be found.

\hfill \mbox{\textit{AQA FP3 2009 Q4 [5]}}
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