OCR FP2 2014 June — Question 1 3 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeStandard integral of 1/√(x²+a²)
DifficultyStandard +0.3 This is a standard Further Maths integral requiring recognition of the inverse hyperbolic sine form (or sinh substitution), followed by straightforward evaluation at limits. While it requires knowledge beyond A-level Core, it's a textbook application of a formula with no problem-solving insight needed, making it slightly easier than average overall.
Spec1.08h Integration by substitution4.07e Inverse hyperbolic: definitions, domains, ranges
1 Find \(\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x\), giving your answer exactly in logarithmic form.
AnswerMarks Guidance
\(\int_0^2 \frac{1}{\sqrt{4+x^2}} dx = \left[\sinh^{-1}\left(\frac{x}{2}\right)\right]_0^2 = \sinh^{-1}1 - \sinh^{-1}0 = \ln 1+\sqrt{1+1}-0 = \ln 1+\sqrt{2}\) cao iswM1, M1, A1 Standard form; Use of log form and substitute limits dep on 1st M
Alternative:
AnswerMarks Guidance
\(\int_0^2 \frac{1}{\sqrt{4+x^2}} dx = \left[\ln x + \sqrt{x^2+4}\right]_0^2 = \ln 2 + \sqrt{8} - \ln 2 = \ln 1+\sqrt{2}\)M1, M1, A1 Standard form; Substitute limits 
$\int_0^2 \frac{1}{\sqrt{4+x^2}} dx = \left[\sinh^{-1}\left(\frac{x}{2}\right)\right]_0^2 = \sinh^{-1}1 - \sinh^{-1}0 = \ln 1+\sqrt{1+1}-0 = \ln 1+\sqrt{2}$ cao isw | M1, M1, A1 | Standard form; Use of log form and substitute limits dep on 1st M

**Alternative:**
$\int_0^2 \frac{1}{\sqrt{4+x^2}} dx = \left[\ln x + \sqrt{x^2+4}\right]_0^2 = \ln 2 + \sqrt{8} - \ln 2 = \ln 1+\sqrt{2}$ | M1, M1, A1 | Standard form; Substitute limits

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1 Find $\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x$, giving your answer exactly in logarithmic form.

\hfill \mbox{\textit{OCR FP2 2014 Q1 [3]}}
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