Edexcel FP3 2013 June — Question 2 5 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Year2013
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeStandard integral of 1/√(x²+a²)
DifficultyStandard +0.3 This is a standard FP3 inverse hyperbolic/trig integration question requiring recognition of the arcsinh form and substitution to match the standard result. Part (a) is direct application of a formula, part (b) requires careful evaluation at limits and simplification of logarithms. While it's Further Maths content (inherently harder), it's a textbook exercise testing recall and technique rather than problem-solving, making it slightly easier than average overall.
Spec1.08d Evaluate definite integrals: between limits4.08h Integration: inverse trig/hyperbolic substitutions
2. (a) Find $$\int \frac { 1 } { \sqrt { } \left( 4 x ^ { 2 } + 9 \right) } d x$$ (b) Use your answer to part (a) to find the exact value of $$\int _ { - 3 } ^ { 3 } \frac { 1 } { \sqrt { \left( 4 x ^ { 2 } + 9 \right) } } d x$$ giving your answer in the form \(k \ln ( a + b \sqrt { } 5 )\), where \(a\) and \(b\) are integers and \(k\) is a constant.
Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(k\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)\) or \(k\ln\!\left[px + \sqrt{p^2x^2 + \frac{9}{4}p^2}\right](+c)\)M1
\(\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)\) or \(\frac{1}{2}\ln\!\left[px + \sqrt{p^2x^2 + \frac{9}{4}p^2}\right](+c)\)A1
(2 marks)
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}\ln\!\left[6+\sqrt{45}\right] - \frac{1}{2}\ln\!\left[-6+\sqrt{45}\right] = \frac{1}{2}\ln\!\left[\frac{6+\sqrt{45}}{-6+\sqrt{45}}\right]\)M1 Uses correct limits and combines logs
\(= \frac{1}{2}\ln\!\left[\frac{6+\sqrt{45}}{-6+\sqrt{45}}\cdot\frac{6+\sqrt{45}}{6+\sqrt{45}}\right] = \frac{1}{2}\ln\!\left[\frac{(6+\sqrt{45})^2}{9}\right]\)M1 Correct method to rationalise denominator (may be implied); method must be clear if answer does not follow their fraction
\(= \ln[2+\sqrt{5}]\) (or \(\frac{1}{2}\ln[9+4\sqrt{5}]\))A1cso
Note: Last 3 marks can also be scored via: \(2\times\frac{1}{2}\Big[\ln[2x+\sqrt{(4x^2+9)}]\Big]_0^3 = \ln(6+\sqrt{45})-\ln 3 = \ln\!\left(\frac{6+\sqrt{45}}{3}\right)\); M1: uses limits 0 and 3 and doubles; M1: combines logs; A1: \(\ln[2+\sqrt{5}]\) oe
(3 marks) — Total: 5
Alternative (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x = \frac{3}{2}\sinh u \Rightarrow \int\frac{1}{\sqrt{9\sinh^2 u+9}}\cdot\frac{3}{2}\cosh u\,du = k\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)\)M1
\(\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)\)A1
Alternative (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\left[\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)\right]_{-3}^{3} = \frac{1}{2}\,\text{arsinh}\;2 - \frac{1}{2}\,\text{arsinh}\;{-2}\)
\(\frac{1}{2}\ln(2+\sqrt{5}) - \frac{1}{2}\ln(\sqrt{5}-2) = \frac{1}{2}\ln\!\left(\frac{2+\sqrt{5}}{\sqrt{5}-2}\right)\)M1 Uses correct limits and combines logs
\(= \frac{1}{2}\ln\frac{2+\sqrt{5}}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{1}{2}\ln\!\left(\frac{2\sqrt{5}+4+5+2\sqrt{5}}{5-4}\right)\)M1 Correct method to rationalise denominator (may be implied)
\(= \frac{1}{2}\ln[9+4\sqrt{5}]\)A1cso 
## Question 2:

**Part (a)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $k\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)$ or $k\ln\!\left[px + \sqrt{p^2x^2 + \frac{9}{4}p^2}\right](+c)$ | M1 | |
| $\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)$ or $\frac{1}{2}\ln\!\left[px + \sqrt{p^2x^2 + \frac{9}{4}p^2}\right](+c)$ | A1 | |

**(2 marks)**

**Part (b)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\ln\!\left[6+\sqrt{45}\right] - \frac{1}{2}\ln\!\left[-6+\sqrt{45}\right] = \frac{1}{2}\ln\!\left[\frac{6+\sqrt{45}}{-6+\sqrt{45}}\right]$ | M1 | Uses correct limits **and** combines logs |
| $= \frac{1}{2}\ln\!\left[\frac{6+\sqrt{45}}{-6+\sqrt{45}}\cdot\frac{6+\sqrt{45}}{6+\sqrt{45}}\right] = \frac{1}{2}\ln\!\left[\frac{(6+\sqrt{45})^2}{9}\right]$ | M1 | Correct method to rationalise denominator (may be implied); method must be clear if answer does not follow their fraction |
| $= \ln[2+\sqrt{5}]$ (or $\frac{1}{2}\ln[9+4\sqrt{5}]$) | A1cso | |

Note: Last 3 marks can also be scored via: $2\times\frac{1}{2}\Big[\ln[2x+\sqrt{(4x^2+9)}]\Big]_0^3 = \ln(6+\sqrt{45})-\ln 3 = \ln\!\left(\frac{6+\sqrt{45}}{3}\right)$; M1: uses limits 0 and 3 and doubles; M1: combines logs; A1: $\ln[2+\sqrt{5}]$ oe

**(3 marks) — Total: 5**

**Alternative (a)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{3}{2}\sinh u \Rightarrow \int\frac{1}{\sqrt{9\sinh^2 u+9}}\cdot\frac{3}{2}\cosh u\,du = k\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)$ | M1 | |
| $\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)(+c)$ | A1 | |

**Alternative (b)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{1}{2}\,\text{arsinh}\!\left(\frac{2x}{3}\right)\right]_{-3}^{3} = \frac{1}{2}\,\text{arsinh}\;2 - \frac{1}{2}\,\text{arsinh}\;{-2}$ | | |
| $\frac{1}{2}\ln(2+\sqrt{5}) - \frac{1}{2}\ln(\sqrt{5}-2) = \frac{1}{2}\ln\!\left(\frac{2+\sqrt{5}}{\sqrt{5}-2}\right)$ | M1 | Uses correct limits **and** combines logs |
| $= \frac{1}{2}\ln\frac{2+\sqrt{5}}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{1}{2}\ln\!\left(\frac{2\sqrt{5}+4+5+2\sqrt{5}}{5-4}\right)$ | M1 | Correct method to rationalise denominator (may be implied) |
| $= \frac{1}{2}\ln[9+4\sqrt{5}]$ | A1cso | |

---
2. (a) Find

$$\int \frac { 1 } { \sqrt { } \left( 4 x ^ { 2 } + 9 \right) } d x$$

(b) Use your answer to part (a) to find the exact value of

$$\int _ { - 3 } ^ { 3 } \frac { 1 } { \sqrt { \left( 4 x ^ { 2 } + 9 \right) } } d x$$

giving your answer in the form $k \ln ( a + b \sqrt { } 5 )$, where $a$ and $b$ are integers and $k$ is a constant.\\

\hfill \mbox{\textit{Edexcel FP3 2013 Q2 [5]}}
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