| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Challenging +1.2 Part (a) is a straightforward exponential integration requiring basic manipulation. Part (b) requires executing a standard substitution u = tan x (with du = sec²x dx), recognizing that sec²x = 1 + tan²x = 1 + u², and evaluating a polynomial integral with fractional powers. While this involves multiple steps and careful algebraic manipulation, it's a textbook application of a given substitution with no novel insight required—slightly above average difficulty due to the algebraic complexity and the need to handle fractional powers correctly. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int_0^{\ln 2} e^{1-2x}\, dx = \left[-\frac{1}{2}e^{1-2x}\right]_0^{\ln 2}\) | M1 A1 | Correct integration |
| \(= -\frac{1}{2}e^{1-2\ln 2} + \frac{1}{2}e^1 = -\frac{1}{2}\cdot\frac{e}{4} + \frac{e}{2}\) | M1 | Substitute limits |
| \(= \frac{e}{2} - \frac{e}{8} = \frac{3e}{8}\) | A1 | Correct simplification |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u = \tan x \Rightarrow \frac{du}{dx} = \sec^2 x\) | B1 | Correct derivative |
| \(\sec^4 x\sqrt{\tan x} = \sec^2 x \cdot \sqrt{u} \cdot \sec^2 x\) | M1 | Attempt substitution |
| \(= (1+u^2)\sqrt{u}\, du\) | A1 | Correct form after substitution |
| Limits: \(x=0 \to u=0\); \(x=\frac{\pi}{4} \to u=1\) | B1 | Both limits changed |
| \(\int_0^1 (u^{1/2} + u^{5/2})\, du\) | A1 | Correct integrand |
| \(= \left[\frac{2}{3}u^{3/2} + \frac{2}{7}u^{7/2}\right]_0^1\) | M1 A1 | Correct integration |
| \(= \frac{2}{3} + \frac{2}{7} = \frac{20}{21}\) | A1 | Correct final answer |
## Question 8:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_0^{\ln 2} e^{1-2x}\, dx = \left[-\frac{1}{2}e^{1-2x}\right]_0^{\ln 2}$ | M1 A1 | Correct integration |
| $= -\frac{1}{2}e^{1-2\ln 2} + \frac{1}{2}e^1 = -\frac{1}{2}\cdot\frac{e}{4} + \frac{e}{2}$ | M1 | Substitute limits |
| $= \frac{e}{2} - \frac{e}{8} = \frac{3e}{8}$ | A1 | Correct simplification |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u = \tan x \Rightarrow \frac{du}{dx} = \sec^2 x$ | B1 | Correct derivative |
| $\sec^4 x\sqrt{\tan x} = \sec^2 x \cdot \sqrt{u} \cdot \sec^2 x$ | M1 | Attempt substitution |
| $= (1+u^2)\sqrt{u}\, du$ | A1 | Correct form after substitution |
| Limits: $x=0 \to u=0$; $x=\frac{\pi}{4} \to u=1$ | B1 | Both limits changed |
| $\int_0^1 (u^{1/2} + u^{5/2})\, du$ | A1 | Correct integrand |
| $= \left[\frac{2}{3}u^{3/2} + \frac{2}{7}u^{7/2}\right]_0^1$ | M1 A1 | Correct integration |
| $= \frac{2}{3} + \frac{2}{7} = \frac{20}{21}$ | A1 | Correct final answer |8
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\int _ { 0 } ^ { \ln 2 } \mathrm { e } ^ { 1 - 2 x } \mathrm {~d} x = \frac { 3 } { 8 } \mathrm { e }$$
\item Use the substitution $u = \tan x$ to find the exact value of
$$\int _ { 0 } ^ { \frac { \pi } { 4 } } \sec ^ { 4 } x \sqrt { \tan x } d x$$
(8 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2013 Q8 [12]}}