| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show definite integral equals specific value (algebraic/exponential substitution) |
| Difficulty | Standard +0.3 This is a straightforward C3 integration question testing standard techniques: recognizing a derivative in the numerator for logarithmic integration (part a) and executing a given substitution with definite integral evaluation (part b). All steps are routine and clearly signposted, making it slightly easier than the average A-level question which typically requires more independent problem-solving. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f' = \frac{dy}{dx} = 4x^3 + 2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{2x^3+1}{x^4+2x}\,dx\) | ||
| \(= \frac{1}{2}\ln(x^4+2x) (+c)\) | M1 A1 | For \(k\ln(x^4+2x)\); by substitution \(k\ln u\) M1, correct A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 2x+1\), \(du = 2\,dx\) | B1 | |
| \(\int x\sqrt{2x+1}\,dx = \int \left(\frac{u-1}{2}\right)\sqrt{u}\,\frac{du}{2}\) | M1 | Must be in terms of \(u\) only incl. \(du\) |
| \(= \frac{1}{4}\int \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right)du\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_0^4 dx = \int_1^9 du\) | B1 | Or changing \(u\)'s to \(x\)'s at end |
| \(\frac{1}{4}\int u^{\frac{3}{2}} - u^{\frac{1}{2}} = \frac{1}{4}\left[\frac{u^{\frac{5}{2}}}{\frac{5}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]\) | M1 A1 | |
| \(= \frac{1}{4}\left[\left(\frac{2}{5}(9)^{\frac{5}{2}} - \frac{2}{3}(9)^{\frac{3}{2}}\right) - \left(\frac{2}{5} - \frac{2}{3}\right)\right]\) | Sight of any of these 3 lines | |
| \(= \frac{1}{4}\left[79.2 + 0.2\dot{6}\right]\) | ||
| \(= 19.86\) | ||
| \(= 19.9\) | A1 | AG |
## Question 3:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f' = \frac{dy}{dx} = 4x^3 + 2$ | B1 | |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{2x^3+1}{x^4+2x}\,dx$ | | |
| $= \frac{1}{2}\ln(x^4+2x) (+c)$ | M1 A1 | For $k\ln(x^4+2x)$; by substitution $k\ln u$ M1, correct A1 |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 2x+1$, $du = 2\,dx$ | B1 | |
| $\int x\sqrt{2x+1}\,dx = \int \left(\frac{u-1}{2}\right)\sqrt{u}\,\frac{du}{2}$ | M1 | Must be in terms of $u$ only incl. $du$ |
| $= \frac{1}{4}\int \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right)du$ | A1 | AG |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^4 dx = \int_1^9 du$ | B1 | Or changing $u$'s to $x$'s at end |
| $\frac{1}{4}\int u^{\frac{3}{2}} - u^{\frac{1}{2}} = \frac{1}{4}\left[\frac{u^{\frac{5}{2}}}{\frac{5}{2}} - \frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]$ | M1 A1 | |
| $= \frac{1}{4}\left[\left(\frac{2}{5}(9)^{\frac{5}{2}} - \frac{2}{3}(9)^{\frac{3}{2}}\right) - \left(\frac{2}{5} - \frac{2}{3}\right)\right]$ | | Sight of any of these 3 lines |
| $= \frac{1}{4}\left[79.2 + 0.2\dot{6}\right]$ | | |
| $= 19.86$ | | |
| $= 19.9$ | A1 | AG |
---3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { f } ( x ) = x ^ { 4 } + 2 x$, find $\mathrm { f } ^ { \prime } ( x )$.
\item Hence, or otherwise, find $\int \frac { 2 x ^ { 3 } + 1 } { x ^ { 4 } + 2 x } \mathrm {~d} x$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the substitution $u = 2 x + 1$ to show that
$$\int x \sqrt { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } \int \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$
\item Hence show that $\int _ { 0 } ^ { 4 } x \sqrt { 2 x + 1 } \mathrm {~d} x = 19.9$ correct to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q3 [10]}}