WJEC Unit 3 Specimen — Question 9 6 marks

Exam BoardWJEC
ModuleUnit 3 (Unit 3)
SessionSpecimen
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeRegion bounded by two curves
DifficultyStandard +0.3 This is a straightforward integration problem requiring finding intersection points (solving x² + 4 = 12 - x², giving x = ±2), then integrating the difference of functions. While it involves multiple steps (finding intersections, setting up integral, integrating, evaluating), all techniques are standard and the problem follows a familiar template for area between curves, making it slightly easier than average.
Spec1.08f Area between two curves: using integration
\includegraphics{figure_9} The diagram above shows a sketch of the curves \(y = x^2 + 4\) and \(y = 12 - x^2\). Find the area of the region bounded by the two curves. [6]
AnswerMarks Guidance
\(x^2 + 4 = 12 - x^2\)M1 (Equating \(y\)'s)
\(2x^2 = 8\)
\(x = \pm 2\)A1
Area \(= \int_{-2}^{2} [(12 - x^2) - (x^2 + 4)] dx\)M1 (expressing area)
\(= \int_{-2}^{2} (8 - 2x^2) dx\)
\(\left[8x - \frac{2x^3}{3}\right]_{-2}^{2}\)A2 (F.T arithmetic error)
\(= \frac{64}{3}\)A1 (c.a.o)
Alternative mark scheme for the Area:
AnswerMarks Guidance
Area \(= \int_{-2}^{2}(12 - x^2)dx - \int_{-2}^{2}(x^2 + 4)dx\)M1
\(\left[12x - \frac{x^3}{3} - \frac{x^3}{3} - 4x\right]_{-2}^{2}\)A2 (A2 for 4 terms correct, A1 for 2 terms correct)
\(= \frac{64}{3}\)A1 (c.a.o)
Total: [6]
$x^2 + 4 = 12 - x^2$ | M1 | (Equating $y$'s)

$2x^2 = 8$ | |

$x = \pm 2$ | A1 |

Area $= \int_{-2}^{2} [(12 - x^2) - (x^2 + 4)] dx$ | M1 | (expressing area)

$= \int_{-2}^{2} (8 - 2x^2) dx$ | |

$\left[8x - \frac{2x^3}{3}\right]_{-2}^{2}$ | A2 | (F.T arithmetic error)

$= \frac{64}{3}$ | A1 | (c.a.o)

**Alternative mark scheme for the Area:**

Area $= \int_{-2}^{2}(12 - x^2)dx - \int_{-2}^{2}(x^2 + 4)dx$ | M1 |

$\left[12x - \frac{x^3}{3} - \frac{x^3}{3} - 4x\right]_{-2}^{2}$ | A2 | (A2 for 4 terms correct, A1 for 2 terms correct)

$= \frac{64}{3}$ | A1 | (c.a.o)

**Total: [6]**
\includegraphics{figure_9}

The diagram above shows a sketch of the curves $y = x^2 + 4$ and $y = 12 - x^2$.

Find the area of the region bounded by the two curves. [6]

\hfill \mbox{\textit{WJEC Unit 3  Q9 [6]}}
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