Edexcel C4 2006 January — Question 3 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2006
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeDefinite integral with complex substitution requiring algebraic rearrangement
DifficultyStandard +0.3 This is a straightforward substitution question where the substitution is explicitly given. Students need to find dx in terms of du, change limits, and integrate a simple polynomial. While it requires careful algebraic manipulation and is worth 8 marks, it's a standard C4 technique with no conceptual surprises, making it slightly easier than average.
Spec1.08h Integration by substitution

3. Using the substitution \(u ^ { 2 } = 2 x - 1\), or otherwise, find the exact value of $$\int _ { 1 } ^ { 5 } \frac { 3 x } { \sqrt { ( 2 x - 1 ) } } \mathrm { d } x$$ (8)
(8)

Question 3:
AnswerMarks Guidance
Working/AnswerMarks Guidance
Uses substitution to obtain \(x = f(u)\): \(\left[\frac{u^2+1}{2}\right]\)M1
And to obtain \(u\frac{du}{dx} = \text{const.}\) or equiv.M1
Reaches \(\int \frac{3(u^2+1)}{2u} u\, du\) or equivalentA1
Simplifies integrand to \(\int\left(3u^2 + \frac{3}{2}\right)du\) or equiv.M1
Integrates to \(\frac{1}{2}u^3 + \frac{3}{2}u\)M1, A1\(\sqrt{}\) A1\(\sqrt{}\) dependent on all previous Ms
Uses new limits \(3\) and \(1\), substituting and subtracting (or returning to function of \(x\) with old limits)M1
Answer \(= 16\)A1 cso [8]
"By Parts" alternative: Attempt at correct direction by partsM1
\(\left[3x(2x-1)^{\frac{1}{2}}\right] - \left\{\int 3(2x-1)^{\frac{1}{2}}\,dx\right\}\)M1{M1A1}
\(\ldots - (2x-1)^{\frac{3}{2}}\)M1A1\(\sqrt{}\)
Uses limits \(5\) and \(1\) correctly; \([42-26] = 16\)M1A1
## Question 3:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Uses substitution to obtain $x = f(u)$: $\left[\frac{u^2+1}{2}\right]$ | M1 | |
| And to obtain $u\frac{du}{dx} = \text{const.}$ or equiv. | M1 | |
| Reaches $\int \frac{3(u^2+1)}{2u} u\, du$ or equivalent | A1 | |
| Simplifies integrand to $\int\left(3u^2 + \frac{3}{2}\right)du$ or equiv. | M1 | |
| Integrates to $\frac{1}{2}u^3 + \frac{3}{2}u$ | M1, A1$\sqrt{}$ | A1$\sqrt{}$ dependent on all previous Ms |
| Uses new limits $3$ and $1$, substituting and subtracting (or returning to function of $x$ with old limits) | M1 | |
| Answer $= 16$ | A1 | cso **[8]** |
| **"By Parts" alternative:** Attempt at correct direction by parts | M1 | |
| $\left[3x(2x-1)^{\frac{1}{2}}\right] - \left\{\int 3(2x-1)^{\frac{1}{2}}\,dx\right\}$ | M1{M1A1} | |
| $\ldots - (2x-1)^{\frac{3}{2}}$ | M1A1$\sqrt{}$ | |
| Uses limits $5$ and $1$ correctly; $[42-26] = 16$ | M1A1 | |

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3. Using the substitution $u ^ { 2 } = 2 x - 1$, or otherwise, find the exact value of

$$\int _ { 1 } ^ { 5 } \frac { 3 x } { \sqrt { ( 2 x - 1 ) } } \mathrm { d } x$$

(8)\\
(8)\\

\hfill \mbox{\textit{Edexcel C4 2006 Q3 [8]}}