| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Definite integral with complex substitution requiring algebraic rearrangement |
| Difficulty | Standard +0.3 Part (a) is a straightforward application of the mid-ordinate rule requiring only careful arithmetic. Part (b) is a standard substitution integral that requires expressing x in terms of u and expanding before integrating - routine for C3 level with no novel insight needed, though the algebraic manipulation adds slight complexity beyond basic substitution. |
| Spec | 1.08h Integration by substitution1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Strip width \(h = \frac{0.4-0}{4} = 0.1\), midpoints at \(x = 0.05, 0.15, 0.25, 0.35\) | M1 | Four strips, correct midpoints identified |
| \(f(0.05) = \cos\sqrt{1.15}\), \(f(0.15) = \cos\sqrt{1.45}\), \(f(0.25) = \cos\sqrt{1.75}\), \(f(0.35) = \cos\sqrt{2.05}\) | M1 | Correct ordinates evaluated |
| \(= 0.1(0.9391 + 0.8855 + 0.8090 + 0.7193)\) | A1 | Correct values (at least 3 correct) |
| \(\approx 0.335\) | A1 | Final answer to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u = 3x+1 \Rightarrow \frac{du}{dx} = 3\), so \(dx = \frac{du}{3}\) | B1 | Correct differential |
| \(x = \frac{u-1}{3}\) | B1 | Correct expression for \(x\) |
| Limits: \(x=0 \Rightarrow u=1\), \(x=1 \Rightarrow u=4\) | B1 | Both limits correct |
| \(\int_1^4 \frac{u-1}{3}\sqrt{u}\cdot\frac{du}{3} = \frac{1}{9}\int_1^4 (u^{3/2} - u^{1/2})\,du\) | M1 | Fully substituted integral in \(u\) only |
| \(= \frac{1}{9}\left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_1^4\) | A1 | Correct integration |
| \(= \frac{1}{9}\left[\left(\frac{64}{5}-\frac{16}{3}\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right]\) | M1 | Substituting limits |
| \(= \frac{1}{9} \cdot \frac{124}{15} = \frac{124}{135}\) | A1 | Correct exact value |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Strip width $h = \frac{0.4-0}{4} = 0.1$, midpoints at $x = 0.05, 0.15, 0.25, 0.35$ | M1 | Four strips, correct midpoints identified |
| $f(0.05) = \cos\sqrt{1.15}$, $f(0.15) = \cos\sqrt{1.45}$, $f(0.25) = \cos\sqrt{1.75}$, $f(0.35) = \cos\sqrt{2.05}$ | M1 | Correct ordinates evaluated |
| $= 0.1(0.9391 + 0.8855 + 0.8090 + 0.7193)$ | A1 | Correct values (at least 3 correct) |
| $\approx 0.335$ | A1 | Final answer to 3 s.f. |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = 3x+1 \Rightarrow \frac{du}{dx} = 3$, so $dx = \frac{du}{3}$ | B1 | Correct differential |
| $x = \frac{u-1}{3}$ | B1 | Correct expression for $x$ |
| Limits: $x=0 \Rightarrow u=1$, $x=1 \Rightarrow u=4$ | B1 | Both limits correct |
| $\int_1^4 \frac{u-1}{3}\sqrt{u}\cdot\frac{du}{3} = \frac{1}{9}\int_1^4 (u^{3/2} - u^{1/2})\,du$ | M1 | Fully substituted integral in $u$ only |
| $= \frac{1}{9}\left[\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right]_1^4$ | A1 | Correct integration |
| $= \frac{1}{9}\left[\left(\frac{64}{5}-\frac{16}{3}\right)-\left(\frac{2}{5}-\frac{2}{3}\right)\right]$ | M1 | Substituting limits |
| $= \frac{1}{9} \cdot \frac{124}{15} = \frac{124}{135}$ | A1 | Correct exact value |
---6
\begin{enumerate}[label=(\alph*)]
\item Use the mid-ordinate rule with four strips to find an estimate for $\int _ { 0 } ^ { 0.4 } \cos \sqrt { 3 x + 1 } \mathrm {~d} x$, giving your answer to three significant figures.
\item Use the substitution $u = 3 x + 1$ to find the exact value of $\int _ { 0 } ^ { 1 } x \sqrt { 3 x + 1 } \mathrm {~d} x$.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q6 [10]}}