OCR C4 2007 January — Question 2 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2007
SessionJanuary
Marks5
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Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyStandard +0.3 This is a straightforward integration by parts question with a standard function pair (x and ln x). While it requires knowing the integration by parts formula and careful execution, it's a textbook exercise with no conceptual difficulty beyond the standard C4 technique, making it slightly easier than average.
Spec1.08i Integration by parts
Find the exact value of \(\int_1^2 x \ln x \, dx\). [5]
AnswerMarks Guidance
Use parts with \(u = \ln x, dv = x\)M1 & give 1st stage in form \(f(x) + l - \int g(x)(dx)\)
Obtain \(\frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 (dx)\)A1 or \(\frac{1}{2}x^2 \ln x - \int \frac{1}{2}x(dx)\)
\(= \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2\) (+c)A1
Use limits correctlyM1
Exact answer \(2 \ln 2 - \frac{3}{4}\)A1 5 marks 
Use parts with $u = \ln x, dv = x$ | M1 | & give 1st stage in form $f(x) + l - \int g(x)(dx)$

Obtain $\frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 (dx)$ | A1 | or $\frac{1}{2}x^2 \ln x - \int \frac{1}{2}x(dx)$

$= \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2$ (+c) | A1 |

Use limits correctly | M1 |

Exact answer $2 \ln 2 - \frac{3}{4}$ | A1 | 5 marks | AEF ISW |

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Find the exact value of $\int_1^2 x \ln x \, dx$. [5]

\hfill \mbox{\textit{OCR C4 2007 Q2 [5]}}
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