CAIE P1 2010 June — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas Between Curves
TypeCurve with Horizontal Line
DifficultyModerate -0.3 This is a straightforward area-between-curves problem requiring finding intersection points by solving a quadratic equation, then evaluating a single definite integral. The algebra is routine and the setup is standard, making it slightly easier than average but still requiring proper technique.
Spec1.08e Area between curve and x-axis: using definite integrals

4 \includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-2_428_550_1343_794} The diagram shows the curve \(y = 6 x - x ^ { 2 }\) and the line \(y = 5\). Find the area of the shaded region.

Question 4:
\(y = 6x - x^2\)
AnswerMarks Guidance
Meets \(y = 5\) when \(x = 1\) or \(x = 5\)B1 co
Integral \(= 3x^2 - \frac{1}{3}x^3\)M1 A1 attempt to integrate. co
Limits (1 to 5) used \(\rightarrow 30\frac{2}{3}\)DM1 value at top limit – value at lower
Area of rectangle \(= 20\)B1\(\sqrt{}\) co to his \(x\) values
Shaded area \(= 10\frac{2}{3}\)A1 [6] co
*(integral of \(6x - x^2 - 5\): B1, M1, A1, DM1 as above, then "\(-5x\)": B1\(\sqrt{}\), A1)*
## Question 4:

$y = 6x - x^2$

Meets $y = 5$ when $x = 1$ or $x = 5$ | B1 | co

Integral $= 3x^2 - \frac{1}{3}x^3$ | M1 A1 | attempt to integrate. co

Limits (1 to 5) used $\rightarrow 30\frac{2}{3}$ | DM1 | value at top limit – value at lower

Area of rectangle $= 20$ | B1$\sqrt{}$ | co to his $x$ values

Shaded area $= 10\frac{2}{3}$ | A1 [6] | co

*(integral of $6x - x^2 - 5$: B1, M1, A1, DM1 as above, then "$-5x$": B1$\sqrt{}$, A1)*

---
4\\
\includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-2_428_550_1343_794}

The diagram shows the curve $y = 6 x - x ^ { 2 }$ and the line $y = 5$. Find the area of the shaded region.

\hfill \mbox{\textit{CAIE P1 2010 Q4 [6]}}