AQA FP3 2009 January — Question 4 6 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2009
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeImproper integral with parts
DifficultyStandard +0.8 Part (a) is a standard integration by parts exercise (difficulty ~-0.5), but part (b) requires handling an improper integral where ln(x) → -∞ as x → 0⁺, demanding proper limit notation and understanding of convergence. The combination of routine technique with non-trivial limiting process elevates this to moderately challenging.
Spec4.08c Improper integrals: infinite limits or discontinuous integrands

4
  1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\), where \(c\) is an arbitrary constant.
  2. Hence evaluate \(\int _ { 0 } ^ { 1 } \ln x \mathrm {~d} x\), showing the limiting process used.

Question 4:
Part (a):
AnswerMarks Guidance
WorkingMarks Guidance
\(\int \ln x\,dx = x\ln x - \int x\!\left(\frac{1}{x}\right)dx\)M1 Integration by parts
\(= x\ln x - x + c\)A1 Total: 2; CSO AG
Part (b):
AnswerMarks Guidance
WorkingMarks Guidance
\(\int_0^1 \ln x\,dx = \lim_{a\to 0}\int_a^1 \ln x\,dx\)M1 OE
\(= \lim_{a\to 0}\{0 - 1 - [a\ln a - a]\}\)M1 \(F(1) - F(a)\) OE
But \(\lim_{a\to 0} a\ln a = 0\)E1 Accept general form e.g. \(\lim_{a\to 0} a^k \ln a = 0\)
So \(\int_0^1 \ln x\,dx = -1\)A1 Total: 4
## Question 4:

### Part (a):
| Working | Marks | Guidance |
|---------|-------|---------|
| $\int \ln x\,dx = x\ln x - \int x\!\left(\frac{1}{x}\right)dx$ | M1 | Integration by parts |
| $= x\ln x - x + c$ | A1 | **Total: 2**; CSO AG |

### Part (b):
| Working | Marks | Guidance |
|---------|-------|---------|
| $\int_0^1 \ln x\,dx = \lim_{a\to 0}\int_a^1 \ln x\,dx$ | M1 | OE |
| $= \lim_{a\to 0}\{0 - 1 - [a\ln a - a]\}$ | M1 | $F(1) - F(a)$ OE |
| But $\lim_{a\to 0} a\ln a = 0$ | E1 | Accept general form e.g. $\lim_{a\to 0} a^k \ln a = 0$ |
| So $\int_0^1 \ln x\,dx = -1$ | A1 | **Total: 4** |

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4
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to show that $\int \ln x \mathrm {~d} x = x \ln x - x + c$, where $c$ is an arbitrary constant.
\item Hence evaluate $\int _ { 0 } ^ { 1 } \ln x \mathrm {~d} x$, showing the limiting process used.
\end{enumerate}

\hfill \mbox{\textit{AQA FP3 2009 Q4 [6]}}