| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Square root substitution: indefinite integral |
| Difficulty | Standard +0.3 This is a slightly-above-average C3 integration question. Part (a) uses a standard 'reverse chain rule' technique where recognizing the derivative in the numerator makes the integral straightforward (ln of denominator). Part (b) is a routine substitution requiring algebraic manipulation and integration of powers. Both are textbook-style exercises with clear signposting, requiring competent technique but no novel insight. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2x^2 - 8x + 3\) | B1 | |
| \(\left(\frac{dy}{dx}\right) = 4x - 8\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_4^6 \frac{x-2}{2x^2 - 8x + 3} dx\) | M1 A1 | M1 for \(k \ln(2x^2 - 8x + 3)\); allow \(k \ln u\) |
| \(= \frac{1}{4} [\ln | 2x^2 - 8x + 3 | ]_4^6\) |
| \(= \frac{1}{4}[\ln 27 - \ln 3]\) | ||
| \(= \frac{1}{4} \ln 9\) | ||
| \(= \frac{1}{2} \ln 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x\sqrt{(3x-1)} dx\) | B1 | OE |
| \(u = 3x - 1\) \(du = 3dx\) | ||
| \(\int = \left(\frac{1}{9}\right) \int (u^{\frac{1}{2}} + u^{-\frac{1}{2}})(du)\) | M1 | 2 terms in \(u\) with rational indices |
| \(= \left(\frac{1}{9}\right) \left[\frac{u^{\frac{5}{2}}}{5} + \frac{u^{\frac{3}{2}}}{3} (+c)\right] / 2 / 2\) | A1 F | Must be 2 terms with correct indices (only ft for \(x = \frac{u-1}{3}\)) |
| \(= \frac{2}{45}(3x-1)^{\frac{5}{2}} + \frac{2}{27}(3x-1)^{\frac{3}{2}} + c\) | A1 | CSO OE |
### 5(a)(i)
$y = 2x^2 - 8x + 3$ | B1 | |
$\left(\frac{dy}{dx}\right) = 4x - 8$ | | | **Total: 1**
### 5(a)(ii)
$\int_4^6 \frac{x-2}{2x^2 - 8x + 3} dx$ | M1 A1 | M1 for $k \ln(2x^2 - 8x + 3)$; allow $k \ln u$ |
$= \frac{1}{4} [\ln|2x^2 - 8x + 3|]_4^6$ | m1 | Correct substitution into $k \ln(2x^2 - 8x + 3)$ or 3, 27 into $k \ln u$ |
$= \frac{1}{4}[\ln 27 - \ln 3]$ | | |
$= \frac{1}{4} \ln 9$ | | |
$= \frac{1}{2} \ln 3$ | A1 | | **Total: 4**
### 5(b)
$\int x\sqrt{(3x-1)} dx$ | B1 | OE |
$u = 3x - 1$ $du = 3dx$ | | |
$\int = \left(\frac{1}{9}\right) \int (u^{\frac{1}{2}} + u^{-\frac{1}{2}})(du)$ | M1 | 2 terms in $u$ with rational indices |
$= \left(\frac{1}{9}\right) \left[\frac{u^{\frac{5}{2}}}{5} + \frac{u^{\frac{3}{2}}}{3} (+c)\right] / 2 / 2$ | A1 F | Must be 2 terms with correct indices (only ft for $x = \frac{u-1}{3}$) |
$= \frac{2}{45}(3x-1)^{\frac{5}{2}} + \frac{2}{27}(3x-1)^{\frac{3}{2}} + c$ | A1 | CSO OE | **Total: 4**5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = 2 x ^ { 2 } - 8 x + 3$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Hence, or otherwise, find
$$\int _ { 4 } ^ { 6 } \frac { x - 2 } { 2 x ^ { 2 } - 8 x + 3 } d x$$
giving your answer in the form $k \ln 3$, where $k$ is a rational number.
\end{enumerate}\item Use the substitution $u = 3 x - 1$ to find $\int x \sqrt { 3 x - 1 } \mathrm {~d} x$, giving your answer in terms of $x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q5 [9]}}