Edexcel C4 2013 June — Question 3 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeDefinite integral with complex substitution requiring algebraic rearrangement
DifficultyStandard +0.8 This is a non-trivial integration by substitution requiring careful manipulation of the given substitution u = 2 + √(2x+1), finding du/dx, changing limits, and simplifying to reach a specific exact form. While the substitution is provided, executing it correctly and obtaining the logarithmic form requires solid algebraic skill and is more demanding than routine C4 integration questions.
Spec1.08h Integration by substitution
3. Using the substitution \(u = 2 + \sqrt { } ( 2 x + 1 )\), or other suitable substitutions, find the exact value of $$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { } ( 2 x + 1 ) } d x$$ giving your answer in the form \(A + 2 \ln B\), where \(A\) is an integer and \(B\) is a positive constant.
Question 3:
\(\int_0^4 \frac{1}{2+\sqrt{(2x+1)}}\,dx\), \(u = 2 + \sqrt{(2x+1)}\)
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\frac{du}{dx} = (2x+1)^{-\frac{1}{2}}\) or \(\frac{dx}{du} = u-2\)M1 Either \(\frac{du}{dx} = \pm K(2x+1)^{-\frac{1}{2}}\) or \(\frac{dx}{du} = \pm\lambda(u-2)\)
\(\frac{du}{dx} = (2x+1)^{-\frac{1}{2}}\) or \(\frac{dx}{du} = (u-2)\)A1 Correct expression
\(\left\{\int\frac{1}{2+\sqrt{(2x+1)}}\,dx\right\} = \int\frac{1}{u}(u-2)\,du\)A1 Correct substitution (ignore integral sign and \(du\))
\(= \int\left(1 - \frac{2}{u}\right)du\)dM1 Attempt to divide each term by \(u\)
\(= u - 2\ln u\)ddM1, A1 ft \(\pm Au \pm B\ln u\); then \(u - 2\ln u\)
\(\left[u - 2\ln u\right]_3^5 = (5-2\ln 5)-(3-2\ln 3)\)M1 Applies limits of 5 and 3 in \(u\) (or 4 and 0 in \(x\)), subtracts correct way
\(= 2 + 2\ln\left(\frac{3}{5}\right)\)A1 cao cso
Total: [8]
# Question 3:

## $\int_0^4 \frac{1}{2+\sqrt{(2x+1)}}\,dx$, $u = 2 + \sqrt{(2x+1)}$

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{du}{dx} = (2x+1)^{-\frac{1}{2}}$ or $\frac{dx}{du} = u-2$ | M1 | Either $\frac{du}{dx} = \pm K(2x+1)^{-\frac{1}{2}}$ or $\frac{dx}{du} = \pm\lambda(u-2)$ |
| $\frac{du}{dx} = (2x+1)^{-\frac{1}{2}}$ or $\frac{dx}{du} = (u-2)$ | A1 | Correct expression |
| $\left\{\int\frac{1}{2+\sqrt{(2x+1)}}\,dx\right\} = \int\frac{1}{u}(u-2)\,du$ | A1 | Correct substitution (ignore integral sign and $du$) |
| $= \int\left(1 - \frac{2}{u}\right)du$ | dM1 | Attempt to divide each term by $u$ |
| $= u - 2\ln u$ | ddM1, A1 ft | $\pm Au \pm B\ln u$; then $u - 2\ln u$ |
| $\left[u - 2\ln u\right]_3^5 = (5-2\ln 5)-(3-2\ln 3)$ | M1 | Applies limits of 5 and 3 in $u$ (or 4 and 0 in $x$), subtracts correct way |
| $= 2 + 2\ln\left(\frac{3}{5}\right)$ | A1 cao cso | |

**Total: [8]**

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3. Using the substitution $u = 2 + \sqrt { } ( 2 x + 1 )$, or other suitable substitutions, find the exact value of

$$\int _ { 0 } ^ { 4 } \frac { 1 } { 2 + \sqrt { } ( 2 x + 1 ) } d x$$

giving your answer in the form $A + 2 \ln B$, where $A$ is an integer and $B$ is a positive constant.\\

\hfill \mbox{\textit{Edexcel C4 2013 Q3 [8]}}
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