CAIE P2 2018 June — Question 3 5 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeDefinite integral with exponentials
DifficultyModerate -0.5 This is a straightforward definite integral requiring expansion of the bracket followed by standard exponential integration. The algebraic manipulation (4e^(-x)(e^(3x)+1) = 4e^(2x) + 4e^(-x)) is routine, and integrating exponentials with the reverse chain rule is a core P2 skill. The exact evaluation at limits 0 and 2 involves simple arithmetic with e^4 and e^(-2), making this slightly easier than average but not trivial.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.08d Evaluate definite integrals: between limits
3 Without using a calculator, find the exact value of \(\int _ { 0 } ^ { 2 } 4 \mathrm { e } ^ { - x } \left( \mathrm { e } ^ { 3 x } + 1 \right) \mathrm { d } x\).
Question 3:
AnswerMarks Guidance
AnswerMark Guidance
Rewrite integrand as \(4e^{2x} + 4e^{-x}\)B1
Integrate to obtain form \(k_1e^{2x} + k_2e^{-x}\) where \(k_1 \neq 4,\, k_2 \neq 4\)M1
Obtain correct \(2e^{2x} - 4e^{-x}\)A1
Apply limits correctly, retaining exactnessM1 Dependent on previous M1
Obtain \(2e^4 - 4e^{-2} + 2\) or exact similarly simplified equivalentA1
Total5 
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| Rewrite integrand as $4e^{2x} + 4e^{-x}$ | B1 | |
| Integrate to obtain form $k_1e^{2x} + k_2e^{-x}$ where $k_1 \neq 4,\, k_2 \neq 4$ | M1 | |
| Obtain correct $2e^{2x} - 4e^{-x}$ | A1 | |
| Apply limits correctly, retaining exactness | M1 | Dependent on previous M1 |
| Obtain $2e^4 - 4e^{-2} + 2$ or exact similarly simplified equivalent | A1 | |
| **Total** | **5** | |

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3 Without using a calculator, find the exact value of $\int _ { 0 } ^ { 2 } 4 \mathrm { e } ^ { - x } \left( \mathrm { e } ^ { 3 x } + 1 \right) \mathrm { d } x$.\\

\hfill \mbox{\textit{CAIE P2 2018 Q3 [5]}}
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Q1 5 Q2 5 Q3 5 Q4 8 Q5 7 Q6 9 Q7 11