Question 1013

Definite Integrals: Areas and Volumes
2013 Paper 1 Question 5 Variant

Question

It is given that
f(x)={1x2a2for axa,0for a<x<4a,f(x) = \begin{cases} \sqrt{ 1 - \frac{x^2}{a^2} } \quad &\textrm{for } -a \leq x \leq a, \\ 0 \quad &\textrm{for } a < x < 4 a, \end{cases}
and that f(x+5a)=f(x){f(x+5 a)=f(x)} for all real values of x,{x,} where a{a} is a real constant.
(i)
Sketch the graph of y=f(x){y=f(x)} for 6ax11a.{- 6 a \leq x \leq 11 a.}
[3]
(ii)
Use the substitution x=acosθ{x=a \cos \theta} to find the exact value of
123a12af(x)  dx\int_{- \frac{1}{2} \sqrt{3} a}^{- \frac{1}{2} a} f(x) \; \mathrm{d}x
in terms of a{a} and π.{\pi.}
[5]

Answer