| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with complex substitution requiring algebraic rearrangement |
| Difficulty | Moderate -0.3 This is a straightforward substitution question where the substitution is given explicitly. Students must change limits, express (5-2x) in terms of u, and integrate a power function. While it requires careful algebraic manipulation and understanding of the substitution method, it follows a standard template with no conceptual surprises, making it slightly easier than average for C3 level. |
| Spec | 1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{du}{dx} = 4\) | B1 | Correct expression for \(\frac{du}{dx}\) or \(du\) or \(dx\) |
| \([u = 4x-1]\), \(4x = u+1\) | B1 | Correct term in \(kx\), where \(k = 1, 2, 4\) |
| \(\int \frac{9-u}{2} \times \sqrt[3]{u} \times \frac{du}{4}\) | M1 | Replacing all terms in \(x\) to all in terms of \(u\), including replacing \(dx\), but condone omission of \(du\) |
| All correct, must see \(du\) | A1 | Must see \(du\) here or on next line |
| \(= \frac{1}{8}\int 9u^{\frac{1}{3}} - u^{\frac{4}{3}} \, du\) | ||
| \(= \frac{1}{8}\left(\frac{3}{4} \times 9u^{\frac{4}{3}} - \frac{3}{7}u^{\frac{7}{3}}\right)\) | m1 | Correct integration from expression of form \(au^{\frac{1}{3}} + bu^{\frac{4}{3}}\) or \(au^{\frac{1}{3}} + bu^{\frac{4}{3}} + cu^{\frac{1}{3}}\) |
| Limits \([x]^{0.5}_{0.25} = [u]^1_0\) | B1 | Or correctly changing variable back into \(x\) |
| \(= \frac{1}{8}\left[\left(\frac{27}{4} - \frac{3}{7}\right) - 0\right]\) | ||
| \(= \frac{177}{224}\) | A1 | Allow equivalent fraction |
# Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{du}{dx} = 4$ | B1 | Correct expression for $\frac{du}{dx}$ or $du$ or $dx$ |
| $[u = 4x-1]$, $4x = u+1$ | B1 | Correct term in $kx$, where $k = 1, 2, 4$ |
| $\int \frac{9-u}{2} \times \sqrt[3]{u} \times \frac{du}{4}$ | M1 | Replacing all terms in $x$ to all in terms of $u$, including replacing $dx$, but condone omission of $du$ |
| All correct, must see $du$ | A1 | Must see $du$ here or on next line |
| $= \frac{1}{8}\int 9u^{\frac{1}{3}} - u^{\frac{4}{3}} \, du$ | | |
| $= \frac{1}{8}\left(\frac{3}{4} \times 9u^{\frac{4}{3}} - \frac{3}{7}u^{\frac{7}{3}}\right)$ | m1 | Correct integration from expression of form $au^{\frac{1}{3}} + bu^{\frac{4}{3}}$ or $au^{\frac{1}{3}} + bu^{\frac{4}{3}} + cu^{\frac{1}{3}}$ |
| Limits $[x]^{0.5}_{0.25} = [u]^1_0$ | B1 | Or correctly changing variable back into $x$ |
| $= \frac{1}{8}\left[\left(\frac{27}{4} - \frac{3}{7}\right) - 0\right]$ | | |
| $= \frac{177}{224}$ | A1 | Allow equivalent fraction |
**Total: 7 marks**
---8 Use the substitution $u = 4 x - 1$ to find the exact value of
$$\int _ { \frac { 1 } { 4 } } ^ { \frac { 1 } { 2 } } ( 5 - 2 x ) ( 4 x - 1 ) ^ { \frac { 1 } { 3 } } \mathrm {~d} x$$
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\includegraphics[max width=\textwidth, alt={}]{bf427498-f1ee-4167-a6f2-ddaa2ff5ef81-18_2104_1712_603_153}
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\hfill \mbox{\textit{AQA C3 2016 Q8 [7]}}