CAIE P3 2011 June — Question 3 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2011
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.3 This is a straightforward 'show that' integration by parts question with a simple polynomial factor (1-x) and exponential function. The definite integral has convenient limits (0 to 1) and requires only one application of integration by parts followed by routine evaluation. Slightly easier than average due to the direct nature and verification format.
Spec1.08i Integration by parts

3 Show that \(\int _ { 0 } ^ { 1 } ( 1 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x } \mathrm {~d} x = 4 \mathrm { e } ^ { - \frac { 1 } { 2 } } - 2\).

AnswerMarks Guidance
Attempt integration by parts and reach \(k(1-x)e^{-\frac{x}{2}} \pm k\int e^{-\frac{x}{2}}dx\), or equivalentM1
Obtain \(-2(1-x)e^{-\frac{x}{2}} - 2\int e^{-\frac{x}{2}}dx\), or equivalentA1
Integrate and obtain \(-2(1-x)e^{-\frac{x}{2}} + 4e^{-\frac{x}{2}}\), or equivalentA1
Use limits \(x = 0\) and \(x = 1\), having integrated twiceM1
Obtain the given answer correctlyA1 [5]
Attempt integration by parts and reach $k(1-x)e^{-\frac{x}{2}} \pm k\int e^{-\frac{x}{2}}dx$, or equivalent | M1 |
Obtain $-2(1-x)e^{-\frac{x}{2}} - 2\int e^{-\frac{x}{2}}dx$, or equivalent | A1 |
Integrate and obtain $-2(1-x)e^{-\frac{x}{2}} + 4e^{-\frac{x}{2}}$, or equivalent | A1 |
Use limits $x = 0$ and $x = 1$, having integrated twice | M1 |
Obtain the given answer correctly | A1 | [5]
3 Show that $\int _ { 0 } ^ { 1 } ( 1 - x ) \mathrm { e } ^ { - \frac { 1 } { 2 } x } \mathrm {~d} x = 4 \mathrm { e } ^ { - \frac { 1 } { 2 } } - 2$.

\hfill \mbox{\textit{CAIE P3 2011 Q3 [5]}}