CAIE P3 2020 June — Question 2 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.3 This is a straightforward single application of integration by parts with simple polynomial and exponential functions, followed by evaluation at given limits. The setup is standard (u = 2-x, dv = e^(-2x)dx) with no conceptual challenges, making it slightly easier than an average A-level question which might require multiple techniques or more complex manipulation.
Spec1.08i Integration by parts
2 Find the exact value of \(\int _ { 0 } ^ { 1 } ( 2 - x ) \mathrm { e } ^ { - 2 x } \mathrm {~d} x\).
Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Commence integration and reach \(a(2-x)e^{-2x} + b\int e^{-2x}\,dx\)M1* or equivalent
Obtain \(-\frac{1}{2}(2-x)e^{-2x} - \frac{1}{2}\int e^{-2x}\,dx\)A1 or equivalent
Complete integration and obtain \(-\frac{1}{2}(2-x)e^{-2x} + \frac{1}{4}e^{-2x}\)A1 or equivalent
Use limits correctly, having integrated twiceDM1
Obtain answer \(\frac{1}{4}(3-e^{-2})\)A1 or exact equivalent 
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $a(2-x)e^{-2x} + b\int e^{-2x}\,dx$ | M1* | or equivalent |
| Obtain $-\frac{1}{2}(2-x)e^{-2x} - \frac{1}{2}\int e^{-2x}\,dx$ | A1 | or equivalent |
| Complete integration and obtain $-\frac{1}{2}(2-x)e^{-2x} + \frac{1}{4}e^{-2x}$ | A1 | or equivalent |
| Use limits correctly, having integrated twice | DM1 | |
| Obtain answer $\frac{1}{4}(3-e^{-2})$ | A1 | or exact equivalent |

---
2 Find the exact value of $\int _ { 0 } ^ { 1 } ( 2 - x ) \mathrm { e } ^ { - 2 x } \mathrm {~d} x$.\\

\hfill \mbox{\textit{CAIE P3 2020 Q2 [5]}}
This paper (9 questions)
View full paper
Q1 4 Q2 5 Q3 6 Q4 6 Q5 6 Q6 7 Q7 9 Q8 10 Q9 10