| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Independent multi-part (different techniques) |
| Difficulty | Moderate -0.3 This is a standard C3 integration question with three routine parts: (a) is a textbook integration by parts example, (b) is straightforward substitution with the substitution given, and (c) is volume of revolution about the y-axis requiring rearrangement to x = √(y+9) and standard formula application. All techniques are standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int x\sin x\, dx\), \(u = x\), \(\dfrac{dv}{dx} = \sin x\) | M1 | For differentiating one term and integrating other |
| \(\dfrac{du}{dx} = 1\), \(v = -\cos x\) | ||
| \(\int = -x\cos x - \int -\cos x\, dx\) | m1 | For correctly substituting into parts formula |
| A1 | ||
| \(= -x\cos x + \sin x\, (+c)\) | A1 | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = x^2 + 5\), \(du = 2x\, dx\) | ||
| \(\int = \int \dfrac{1}{2} u^{\frac{1}{2}}\, du\) | M1 | \(\int ku^{\frac{1}{2}}\, du\) condone omission of \(du\); M0 if \(dx\); \(k = \frac{1}{2}\) OE |
| A1 | ||
| \(= \dfrac{u^{\frac{3}{2}}}{3}\) | A1\(\checkmark\) | Ft \(\int ku^{\frac{1}{2}}\, du\) |
| \(= \dfrac{1}{3}\sqrt{(x^2+5)^3}\, (+c)\) | A1 | CSO; SC \(\dfrac{2}{6}\sqrt{(x^2+5)^3}\) with no working B3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = x^2 - 9\), \(x^2 = y + 9\) | ||
| \(V = \pi\int x^2\, dy = \pi\int(y+9)\, dy\) | B1 | Must have \(\pi\) and \(x^2\); condone omission of \(dy\), but B0 if \(dx\) |
| \(= (\pi)\left[\dfrac{y^2}{2} + 9y\right]_1^2\) or \((\pi)\left[\dfrac{(y+9)^2}{2}\right]_1^2\) | M1 | \(\int\)"their \(x^2\)"\(dy\) integrated; limits 2 and 1 substituted in correct order including sign; \(\pi\) not necessary |
| \(= (\pi)\left[20 - 9\tfrac{1}{2}\right]\) | m1 | |
| \(= 10\tfrac{1}{2}\pi\) | A1 | CSO |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int x\sin x\, dx$, $u = x$, $\dfrac{dv}{dx} = \sin x$ | M1 | For differentiating one term and integrating other |
| $\dfrac{du}{dx} = 1$, $v = -\cos x$ | | |
| $\int = -x\cos x - \int -\cos x\, dx$ | m1 | For correctly substituting into parts formula |
| | A1 | |
| $= -x\cos x + \sin x\, (+c)$ | A1 | CSO |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = x^2 + 5$, $du = 2x\, dx$ | | |
| $\int = \int \dfrac{1}{2} u^{\frac{1}{2}}\, du$ | M1 | $\int ku^{\frac{1}{2}}\, du$ condone omission of $du$; M0 if $dx$; $k = \frac{1}{2}$ OE |
| | A1 | |
| $= \dfrac{u^{\frac{3}{2}}}{3}$ | A1$\checkmark$ | Ft $\int ku^{\frac{1}{2}}\, du$ |
| $= \dfrac{1}{3}\sqrt{(x^2+5)^3}\, (+c)$ | A1 | CSO; SC $\dfrac{2}{6}\sqrt{(x^2+5)^3}$ with no working B3 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = x^2 - 9$, $x^2 = y + 9$ | | |
| $V = \pi\int x^2\, dy = \pi\int(y+9)\, dy$ | B1 | Must have $\pi$ and $x^2$; condone omission of $dy$, but B0 if $dx$ |
| $= (\pi)\left[\dfrac{y^2}{2} + 9y\right]_1^2$ or $(\pi)\left[\dfrac{(y+9)^2}{2}\right]_1^2$ | M1 | $\int$"their $x^2$"$dy$ integrated; limits 2 and 1 substituted in correct order including sign; $\pi$ not necessary |
| $= (\pi)\left[20 - 9\tfrac{1}{2}\right]$ | m1 | |
| $= 10\tfrac{1}{2}\pi$ | A1 | CSO |
**Total: 12 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to find $\int x \sin x \mathrm {~d} x$.
\item Using the substitution $u = x ^ { 2 } + 5$, or otherwise, find $\int x \sqrt { x ^ { 2 } + 5 } \mathrm {~d} x$.
\item The diagram shows the curve $y = x ^ { 2 } - 9$ for $x \geqslant 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-3_844_663_685_694}
The shaded region $R$ is bounded by the curve, the lines $y = 1$ and $y = 2$, and the $y$-axis.
Find the exact value of the volume of the solid generated when the region $R$ is rotated through $360 ^ { \circ }$ about the $\boldsymbol { y }$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q4 [12]}}