Edexcel F3 2017 June — Question 7 8 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2017
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeIntegration with substitution given
DifficultyStandard +0.8 This Further Pure F3 question requires splitting the integral into two parts: a standard arcsin integral and a reverse chain rule component. Part (b) demands careful evaluation of exact values including arcsin(√3/2) = π/3 and algebraic manipulation to reach the specified form. While systematic, it requires confident handling of inverse trig functions and exact values beyond typical A-level Core content.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08d Evaluate definite integrals: between limits1.08h Integration by substitution
  1. (a) Find
$$\int \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } \mathrm {~d} x$$ (b) Hence find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } d x$$ giving your answer in the form \(p \pi \sqrt { 3 } + q\), where \(p\) and \(q\) are rational numbers to be found.
Question 7:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\int \frac{5+x}{\sqrt{4-3x^2}}dx = \int \frac{5}{\sqrt{4-3x^2}}dx + \int \frac{x}{\sqrt{4-3x^2}}dx\)M1 Splits with correct denominators; evidenced by two separate integrals below
\(\int \frac{5}{\sqrt{4-3x^2}}dx = \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x\)dM1A1 M1: \(p\arcsin qx\), depends on 1st M (p can=1); A1: \(\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x\)
\(\int \frac{x}{\sqrt{4-3x^2}}dx = -\frac{1}{3}(4-3x^2)^{\frac{1}{2}}\)dM1A1 M1: \(k(4-3x^2)^{\frac{1}{2}}\), depends on 1st M; A1: \(-\frac{1}{3}(4-3x^2)^{\frac{1}{2}}\)
\(\int \frac{5+x}{\sqrt{4-3x^2}}dx = \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}(+c)\) Total: (5)
Alternative (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x=\frac{2}{\sqrt{3}}\sin u \Rightarrow \int \frac{5+x}{\sqrt{4-3x^2}}dx = \int \frac{5+\frac{2}{\sqrt{3}}\sin u}{\sqrt{4-4\sin^2 u}}\cdot\frac{2}{\sqrt{3}}\cos u\, du\)M1 Attempt \(x=k\sin u\) including replacing \(dx\)
\(= \frac{5}{\sqrt{3}}u - \frac{2}{3}\cos u (+c)\)dM1A1 M1: \(ku\) or \(k\cos u\), depends on 1st M; A1: Both correct
\(= \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}(+c)\)ddM1A1 M1: Changes back to \(x\), depends on both preceding M marks; A1: Fully correct (allow equivalent correct forms)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\left[\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}\right]_0^1\)
\(\left[\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2} - \frac{1}{3}\right] - \left[\frac{5}{\sqrt{3}}\arcsin 0 - \frac{1}{3}\times 2\right]\)M1 Substitute limits 0 and 1 (or 0 and \(\frac{\pi}{3}\) if in terms of \(u\)) in both parts of integral from (a) and subtract the right way round
\(= \frac{5}{9}\pi\sqrt{3} + \frac{1}{3}\)A1, A1 Any exact equivalent; Total: (3) 
## Question 7:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{5+x}{\sqrt{4-3x^2}}dx = \int \frac{5}{\sqrt{4-3x^2}}dx + \int \frac{x}{\sqrt{4-3x^2}}dx$ | M1 | Splits with correct denominators; evidenced by two separate integrals below |
| $\int \frac{5}{\sqrt{4-3x^2}}dx = \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x$ | dM1A1 | M1: $p\arcsin qx$, depends on 1st M (p can=1); A1: $\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x$ |
| $\int \frac{x}{\sqrt{4-3x^2}}dx = -\frac{1}{3}(4-3x^2)^{\frac{1}{2}}$ | dM1A1 | M1: $k(4-3x^2)^{\frac{1}{2}}$, depends on 1st M; A1: $-\frac{1}{3}(4-3x^2)^{\frac{1}{2}}$ |
| $\int \frac{5+x}{\sqrt{4-3x^2}}dx = \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}(+c)$ | | Total: (5) |

**Alternative (a):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=\frac{2}{\sqrt{3}}\sin u \Rightarrow \int \frac{5+x}{\sqrt{4-3x^2}}dx = \int \frac{5+\frac{2}{\sqrt{3}}\sin u}{\sqrt{4-4\sin^2 u}}\cdot\frac{2}{\sqrt{3}}\cos u\, du$ | M1 | Attempt $x=k\sin u$ including replacing $dx$ |
| $= \frac{5}{\sqrt{3}}u - \frac{2}{3}\cos u (+c)$ | dM1A1 | M1: $ku$ or $k\cos u$, depends on 1st M; A1: Both correct |
| $= \frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}(+c)$ | ddM1A1 | M1: Changes back to $x$, depends on both preceding M marks; A1: Fully correct (allow equivalent correct forms) |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2}x - \frac{1}{3}(4-3x^2)^{\frac{1}{2}}\right]_0^1$ | | |
| $\left[\frac{5}{\sqrt{3}}\arcsin\frac{\sqrt{3}}{2} - \frac{1}{3}\right] - \left[\frac{5}{\sqrt{3}}\arcsin 0 - \frac{1}{3}\times 2\right]$ | M1 | Substitute limits 0 and 1 (or 0 and $\frac{\pi}{3}$ if in terms of $u$) in both parts of integral from (a) and subtract the right way round |
| $= \frac{5}{9}\pi\sqrt{3} + \frac{1}{3}$ | A1, A1 | Any exact equivalent; Total: (3) |

---
\begin{enumerate}
  \item (a) Find
\end{enumerate}

$$\int \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } \mathrm {~d} x$$

(b) Hence find the exact value of

$$\int _ { 0 } ^ { 1 } \frac { 5 + x } { \sqrt { 4 - 3 x ^ { 2 } } } d x$$

giving your answer in the form $p \pi \sqrt { 3 } + q$, where $p$ and $q$ are rational numbers to be found.\\

\hfill \mbox{\textit{Edexcel F3 2017 Q7 [8]}}
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