OCR C4 2007 January — Question 4 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2007
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow definite integral equals specific value (algebraic/exponential substitution)
DifficultyModerate -0.8 This is a straightforward integration by substitution question with clear guidance. The substitution is given, requiring only mechanical application: find du = 2dx so 4x-8 = 2(2x-5)+2 = 2u+2, then integrate (2u+2)u^7 which is routine polynomial integration. The algebra is simple and the method is standard textbook practice, making it easier than average.
Spec1.08h Integration by substitution

Use the substitution \(u = 2x - 5\) to show that \(\int_2^3 (4x - 8)(2x - 5)^7 \, dx = \frac{17}{72}\). [5]

AnswerMarks Guidance
Attempt to connect \(dx\) and \(du\)M1 but not just \(dx = du\)
For \(du = 2 dx\) AEF correctly usedA1 sight of \(\frac{1}{2}(du)\) necessary
\(\int u^3 + u^7 (du)\)A1 or \(\int u^7(u+1)(du)\)
Attempt new limits for \(u\) at any stage (expect 0,1)M1 or re-substitute & use \((\frac{3}{2}, 3)\)
\(\frac{12}{7}\)A1 5 marks
S.R. If M1 A0 A0 M1 A0, award S.R. B1 for answer \(\frac{68}{72}, \frac{34}{36}\) or \(\frac{17}{18}\)ISW
Attempt to connect $dx$ and $du$ | M1 | but not just $dx = du$

For $du = 2 dx$ AEF correctly used | A1 | sight of $\frac{1}{2}(du)$ necessary

$\int u^3 + u^7 (du)$ | A1 | or $\int u^7(u+1)(du)$

Attempt new limits for $u$ at any stage (expect 0,1) | M1 | or re-substitute & use $(\frac{3}{2}, 3)$

$\frac{12}{7}$ | A1 | 5 marks | AG WWW |

S.R. If M1 A0 A0 M1 A0, award S.R. B1 for answer $\frac{68}{72}, \frac{34}{36}$ or $\frac{17}{18}$ | ISW |

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Use the substitution $u = 2x - 5$ to show that $\int_2^3 (4x - 8)(2x - 5)^7 \, dx = \frac{17}{72}$. [5]

\hfill \mbox{\textit{OCR C4 2007 Q4 [5]}}