CAIE P1 2004 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2004
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeDefinite integral with simple linear/polynomial substitution
DifficultyModerate -0.5 This is a straightforward integration by substitution question requiring the standard technique of letting u = 3x + 1, adjusting limits, and integrating u^(1/2). It's slightly easier than average because it's a single-step substitution with simple arithmetic and no algebraic complications, though it does require proper handling of definite integral limits.
Spec1.08h Integration by substitution

2 Evaluate \(\int _ { 0 } ^ { 1 } \sqrt { } ( 3 x + 1 ) \mathrm { d } x\).

\(\int_0^1 \sqrt{3x + 1}dx = (3x+1)^{5/3} + 1.5\) then \(÷ 3\)
AnswerMarks
B1 M1 M1MI for \((3x+1)^{5/3} + 1.5\); For division by 3
\(\to [ \ ]\) at \(1 - [ \ ]\) at \(0\)
AnswerMarks
M1Must attempt \([ \ ]\) at \(x=0\) (not assume it is 0) and be using an integrated function
\(\to 16/9 - 2/9 = 14/9\) or \(1.56\)
AnswerMarks
A1 [4]Fraction or decimal. (\(1.56+\)C loses this A1)
$\int_0^1 \sqrt{3x + 1}dx = (3x+1)^{5/3} + 1.5$ then $÷ 3$
| B1 M1 M1 | MI for $(3x+1)^{5/3} + 1.5$; For division by 3 |

$\to [ \ ]$ at $1 - [ \ ]$ at $0$
| M1 | Must attempt $[ \ ]$ at $x=0$ (not assume it is 0) and be using an integrated function |

$\to 16/9 - 2/9 = 14/9$ or $1.56$
| A1 [4] | Fraction or decimal. ($1.56+$C loses this A1) |

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2 Evaluate $\int _ { 0 } ^ { 1 } \sqrt { } ( 3 x + 1 ) \mathrm { d } x$.

\hfill \mbox{\textit{CAIE P1 2004 Q2 [4]}}