Edexcel C2 — Question 2 6 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeShow definite integral equals value
DifficultyModerate -0.5 This is a straightforward C2 integration question requiring standard power rule application (rewriting surds as fractional powers), evaluating at limits, and simplifying to match the given form. It's slightly easier than average because it's a 'show that' question with a clear target, involves only routine techniques, and the algebraic simplification is manageable.
Spec1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits
2. Show that $$\int _ { 2 } ^ { 3 } \left( 6 \sqrt { x } - \frac { 4 } { \sqrt { x } } \right) \mathrm { d } x = k \sqrt { 3 } ,$$ where \(k\) is an integer to be found.
AnswerMarks
\(\int_2^3 \left(6\sqrt{x} - \frac{4}{\sqrt{x}}\right) dx = \left[4x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\right]_2^3\)M1 A2
\(= [4(3\sqrt{3}) - 8\sqrt{3}] - [4(2\sqrt{2}) - 8\sqrt{2}]\)M1 B1
\(= (12\sqrt{3} - 8\sqrt{3}) - (8\sqrt{2} - 8\sqrt{2})\)
\(= 4\sqrt{3}\)A1
\([k = 4]\) 
$\int_2^3 \left(6\sqrt{x} - \frac{4}{\sqrt{x}}\right) dx = \left[4x^{\frac{3}{2}} - 8x^{\frac{1}{2}}\right]_2^3$ | M1 A2 |
$= [4(3\sqrt{3}) - 8\sqrt{3}] - [4(2\sqrt{2}) - 8\sqrt{2}]$ | M1 B1 |
$= (12\sqrt{3} - 8\sqrt{3}) - (8\sqrt{2} - 8\sqrt{2})$ | |
$= 4\sqrt{3}$ | A1 |
$[k = 4]$ | |
2. Show that

$$\int _ { 2 } ^ { 3 } \left( 6 \sqrt { x } - \frac { 4 } { \sqrt { x } } \right) \mathrm { d } x = k \sqrt { 3 } ,$$

where $k$ is an integer to be found.\\

\hfill \mbox{\textit{Edexcel C2  Q2 [6]}}
This paper (9 questions)
View full paper
Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 9 Q7 9 Q8 12 Q9 12