OCR C3 2006 January — Question 1 4 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2006
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with logarithmic form
DifficultyEasy -1.2 This is a straightforward application of the standard integral ∫(1/x)dx = ln|x|. It requires only direct substitution into the formula and basic logarithm laws (ln 8 - ln 2 = ln(8/2) = ln 4 = ln 64 is actually ln 4, or recognizing 3(ln 8 - ln 2) = 3 ln 4 = ln 64). This is simpler than average C3 questions as it's purely procedural with no problem-solving element.
Spec1.06f Laws of logarithms: addition, subtraction, power rules1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits

1 Show that \(\int _ { 2 } ^ { 8 } \frac { 3 } { x } \mathrm {~d} x = \ln 64\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Obtain integral of form \(k\ln x\)M1 Any non-zero constant \(k\); or equiv such as \(k\ln 3x\)
Obtain \(3\ln 8 - 3\ln 2\)A1 Or exact equiv
Attempt use of at least one relevant log propertyM1 Would be earned by initial \(\ln x^3\)
Obtain \(3\ln 4\) or \(\ln 8^3 - \ln 2^3\) and hence \(\ln 64\)A1 4 [AG; with no errors]
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain integral of form $k\ln x$ | M1 | Any non-zero constant $k$; or equiv such as $k\ln 3x$ |
| Obtain $3\ln 8 - 3\ln 2$ | A1 | Or exact equiv |
| Attempt use of at least one relevant log property | M1 | Would be earned by initial $\ln x^3$ |
| Obtain $3\ln 4$ or $\ln 8^3 - \ln 2^3$ and hence $\ln 64$ | A1 | **4** [AG; with no errors] |

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1 Show that $\int _ { 2 } ^ { 8 } \frac { 3 } { x } \mathrm {~d} x = \ln 64$.

\hfill \mbox{\textit{OCR C3 2006 Q1 [4]}}