AQA C3 2015 June — Question 7 7 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeDefinite integral with complex substitution requiring algebraic rearrangement
DifficultyStandard +0.3 This is a standard C3 integration by substitution question with clear guidance (substitution given). Students must find du/dx, change limits, rewrite the integrand in terms of u, and integrate. The algebraic manipulation is straightforward and the final form is specified. Slightly above average difficulty due to the multiple steps and need to express the answer in surd form, but remains a routine examination question.
Spec1.08h Integration by substitution

7 Use the substitution \(u = 6 - x ^ { 2 }\) to find the value of \(\int _ { 1 } ^ { 2 } \frac { x ^ { 3 } } { \sqrt { 6 - x ^ { 2 } } } \mathrm {~d} x\), giving your answer in the form \(p \sqrt { 5 } + q \sqrt { 2 }\), where \(p\) and \(q\) are rational numbers.
[0pt] [7 marks]

Question 7:
AnswerMarks Guidance
\(u = 6 - x^2 \Rightarrow \frac{du}{dx} = -2x\), so \(x^2 = 6 - u\)M1 Correct differentiation of substitution
\(x^3\, dx = x^2 \cdot x\, dx = (6-u)\cdot\left(-\frac{1}{2}\right)du\)M1 Expressing \(x^3\, dx\) in terms of \(u\)
Limits: \(x=1 \Rightarrow u=5\); \(x=2 \Rightarrow u=2\)B1 Both limits correct
\(\int_5^2 \frac{(6-u)(-\frac{1}{2})}{\sqrt{u}}\, du = \frac{1}{2}\int_2^5 \frac{6-u}{\sqrt{u}}\, du\)A1 Correct integral in \(u\)
\(= \frac{1}{2}\int_2^5 \left(6u^{-1/2} - u^{1/2}\right)du\)M1 Splitting and preparing to integrate
\(= \frac{1}{2}\left[12u^{1/2} - \frac{2}{3}u^{3/2}\right]_2^5\)A1 Correct integration
\(= \frac{1}{2}\left[\left(12\sqrt{5} - \frac{2}{3} \cdot 5\sqrt{5}\right) - \left(12\sqrt{2} - \frac{2}{3} \cdot 2\sqrt{2}\right)\right] = \frac{22}{3}\sqrt{5} - \frac{16}{3}\sqrt{2}\)A1 \(p = \frac{22}{3}\), \(q = -\frac{16}{3}\)
I can see these are AQA exam paper answer space pages (P/Jun15/MPC3), but they don't contain the mark scheme - they show only blank answer spaces for Questions 7 and 8.
To help you with the mark scheme content for these questions, I can instead work through the solutions based on the questions shown:
# Question 7:
| $u = 6 - x^2 \Rightarrow \frac{du}{dx} = -2x$, so $x^2 = 6 - u$ | M1 | Correct differentiation of substitution |
|---|---|---|
| $x^3\, dx = x^2 \cdot x\, dx = (6-u)\cdot\left(-\frac{1}{2}\right)du$ | M1 | Expressing $x^3\, dx$ in terms of $u$ |
| Limits: $x=1 \Rightarrow u=5$; $x=2 \Rightarrow u=2$ | B1 | Both limits correct |
| $\int_5^2 \frac{(6-u)(-\frac{1}{2})}{\sqrt{u}}\, du = \frac{1}{2}\int_2^5 \frac{6-u}{\sqrt{u}}\, du$ | A1 | Correct integral in $u$ |
| $= \frac{1}{2}\int_2^5 \left(6u^{-1/2} - u^{1/2}\right)du$ | M1 | Splitting and preparing to integrate |
| $= \frac{1}{2}\left[12u^{1/2} - \frac{2}{3}u^{3/2}\right]_2^5$ | A1 | Correct integration |
| $= \frac{1}{2}\left[\left(12\sqrt{5} - \frac{2}{3} \cdot 5\sqrt{5}\right) - \left(12\sqrt{2} - \frac{2}{3} \cdot 2\sqrt{2}\right)\right] = \frac{22}{3}\sqrt{5} - \frac{16}{3}\sqrt{2}$ | A1 | $p = \frac{22}{3}$, $q = -\frac{16}{3}$ |

I can see these are AQA exam paper answer space pages (P/Jun15/MPC3), but they don't contain the mark scheme - they show only blank answer spaces for Questions 7 and 8.

To help you with the mark scheme content for these questions, I can instead **work through the solutions** based on the questions shown:

---
7 Use the substitution $u = 6 - x ^ { 2 }$ to find the value of $\int _ { 1 } ^ { 2 } \frac { x ^ { 3 } } { \sqrt { 6 - x ^ { 2 } } } \mathrm {~d} x$, giving your answer in the form $p \sqrt { 5 } + q \sqrt { 2 }$, where $p$ and $q$ are rational numbers.\\[0pt]
[7 marks]

\hfill \mbox{\textit{AQA C3 2015 Q7 [7]}}