Question 1013

Definite Integrals: Areas and Volumes

2013 Paper 1 Question 5 Variant

Question

It is given that

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 < 3 a, \end{cases}

and that

{f(x+4 a)=f(x)}

for all real values of

{x,}

where

{a}

is a real constant.

(i)

Sketch the graph of

{y=f(x)}

for

{- 7 a \leq x \leq 6 a.}

[3]

(ii)

Use the substitution

{x=a \cos \theta}

to find the exact value of

\int_{- \frac{1}{2} \sqrt{2} a}^{0} f(x) \; \mathrm{d}x

in terms of

{a}

and

{\pi.}

[5]

Answer

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