| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring standard integration techniques: (i) finding mean value of a derivative using the integral formula, and (ii) applying centroid formulas with exponential functions. While it involves Further Maths content and multiple integrations, the methods are direct applications of learned formulas with no conceptual challenges or novel problem-solving required. |
| Spec | 1.06j 4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{M.V.} = \dfrac{\left[e^{-2x}\right]_0^2}{2-0} = \dfrac{e^{-4}-1}{2}\ (= -0.491)\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = \dfrac{\int_0^2 xe^{-2x}\,dx}{\int_0^2 e^{-2x}\,dx}\) and \(\bar{y} = \dfrac{1}{2}\dfrac{\int_0^2 e^{-4x}\,dx}{\int_0^2 e^{-2x}\,dx}\) | M1M1 | |
| \(\int_0^2 e^{-2x}\,dx = \left[-\tfrac{1}{2}e^{-2x}\right]_0^2 = \tfrac{1}{2}(1-e^{-4})\ (= 0.4908\ldots)\) | B1 | N.B. as in (i) |
| \(\int_0^2 xe^{-2x}\,dx = \left[-\tfrac{xe^{-2x}}{2}\right]_0^2 + \int_0^2 \tfrac{e^{-2x}}{2}\,dx = \left[-\tfrac{xe^{-2x}}{2} - \tfrac{e^{-2x}}{4}\right]_0^2 = -\tfrac{5e^{-4}}{4} + \tfrac{1}{4}\ (= 0.2271\ldots)\) | M1A1 | |
| \(\tfrac{1}{2}\int_0^2 e^{-4x}\,dx = \tfrac{1}{2}\left[-\tfrac{e^{-4x}}{4}\right]_0^2 = \tfrac{1}{2}\left(\tfrac{1}{4} - \tfrac{e^{-8}}{4}\right)\ (= 0.12495\ldots)\) | M1A1 | not straight from calculator |
| \(\bar{x} = 0.463\), \(\bar{y} = 0.255\) | A1A1 | Final answers in algebraic form SRA1 |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{M.V.} = \dfrac{\left[e^{-2x}\right]_0^2}{2-0} = \dfrac{e^{-4}-1}{2}\ (= -0.491)$ | M1A1 | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \dfrac{\int_0^2 xe^{-2x}\,dx}{\int_0^2 e^{-2x}\,dx}$ and $\bar{y} = \dfrac{1}{2}\dfrac{\int_0^2 e^{-4x}\,dx}{\int_0^2 e^{-2x}\,dx}$ | M1M1 | |
| $\int_0^2 e^{-2x}\,dx = \left[-\tfrac{1}{2}e^{-2x}\right]_0^2 = \tfrac{1}{2}(1-e^{-4})\ (= 0.4908\ldots)$ | B1 | N.B. as in (i) |
| $\int_0^2 xe^{-2x}\,dx = \left[-\tfrac{xe^{-2x}}{2}\right]_0^2 + \int_0^2 \tfrac{e^{-2x}}{2}\,dx = \left[-\tfrac{xe^{-2x}}{2} - \tfrac{e^{-2x}}{4}\right]_0^2 = -\tfrac{5e^{-4}}{4} + \tfrac{1}{4}\ (= 0.2271\ldots)$ | M1A1 | |
| $\tfrac{1}{2}\int_0^2 e^{-4x}\,dx = \tfrac{1}{2}\left[-\tfrac{e^{-4x}}{4}\right]_0^2 = \tfrac{1}{2}\left(\tfrac{1}{4} - \tfrac{e^{-8}}{4}\right)\ (= 0.12495\ldots)$ | M1A1 | not straight from calculator |
| $\bar{x} = 0.463$, $\bar{y} = 0.255$ | A1A1 | Final answers in algebraic form SR**A1** |
---7 The curve $C$ has equation $y = \mathrm { e } ^ { - 2 x }$. Find, giving your answers correct to 3 significant figures,\\
(i) the mean value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ over the interval $0 \leqslant x \leqslant 2$,\\
(ii) the coordinates of the centroid of the region bounded by $C$, $x = 0$, $x = 2$ and $y = 0$.
\hfill \mbox{\textit{CAIE FP1 2016 Q7 [11]}}