CAIE P1 2008 June — Question 9 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard Integrals and Reverse Chain Rule
TypeFind curve equation from derivative (find unknown constant in derivative first)
DifficultyModerate -0.8 This is a straightforward integration question requiring basic power rule integration (x^{-3} → x^{-2}/-2), finding the constant of integration using given points, and then computing a definite integral for area. All steps are routine A-level techniques with no problem-solving insight required, making it easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals

9 \includegraphics[max width=\textwidth, alt={}, center]{d71002bb-b6f0-42a3-89fb-f2769d5c3779-3_791_885_1281_630} The diagram shows a curve for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { k } { x ^ { 3 } }\), where \(k\) is a constant. The curve passes through the points \(( 1,18 )\) and \(( 4,3 )\).
  1. Show, by integration, that the equation of the curve is \(y = \frac { 16 } { x ^ { 2 } } + 2\). The point \(P\) lies on the curve and has \(x\)-coordinate 1.6.
  2. Find the area of the shaded region.

\(\frac{dy}{dx} = \frac{k}{x^3}\)
AnswerMarks Guidance
(i) Integrating \(\quad y = -k\frac{x^{-2}}{-2} (+ c)\)B1 ok unsimplified
Sub \((1,18)\) \(\quad 18 = \frac{k}{2} + c\)M1 Substitutes once (even if without + c)
Sub \((4,3)\) \(\quad 3 = \frac{k}{32} + c\)M1 2nd substitution and solution of simultaneous equations for \(k\) and \(c\)
\(\rightarrow k = 32, c = 2\)A1 co [4]
(ii) Area \(= \left[-\frac{16}{x} + 2x\right]\) from 1 to 1.6B1 B1 co
\(\rightarrow [-10 + 3.2] - [-16 + 2] = 7.2\)M1 A1 Use of limits in an integral. co. [4]
$\frac{dy}{dx} = \frac{k}{x^3}$

**(i)** Integrating $\quad y = -k\frac{x^{-2}}{-2} (+ c)$ | B1 | ok unsimplified

Sub $(1,18)$ $\quad 18 = \frac{k}{2} + c$ | M1 | Substitutes once (even if without + c)

Sub $(4,3)$ $\quad 3 = \frac{k}{32} + c$ | M1 | 2nd substitution and solution of simultaneous equations for $k$ and $c$

$\rightarrow k = 32, c = 2$ | A1 | co [4]

**(ii)** Area $= \left[-\frac{16}{x} + 2x\right]$ from 1 to 1.6 | B1 B1 | co

$\rightarrow [-10 + 3.2] - [-16 + 2] = 7.2$ | M1 A1 | Use of limits in an integral. co. [4]
9\\
\includegraphics[max width=\textwidth, alt={}, center]{d71002bb-b6f0-42a3-89fb-f2769d5c3779-3_791_885_1281_630}

The diagram shows a curve for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { k } { x ^ { 3 } }$, where $k$ is a constant. The curve passes through the points $( 1,18 )$ and $( 4,3 )$.\\
(i) Show, by integration, that the equation of the curve is $y = \frac { 16 } { x ^ { 2 } } + 2$.

The point $P$ lies on the curve and has $x$-coordinate 1.6.\\
(ii) Find the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2008 Q9 [8]}}