Author Topic: Worked solution/explanation please? (Read 777 times) Tweet Share

srxmos · « **on:** July 10, 2016, 01:57:10 pm »

Calculator allowed too

RuiAce · « **Reply #1 on:** July 10, 2016, 02:06:34 pm »

RuiAce · « **Reply #2 on:** July 10, 2016, 02:09:37 pm »

Quote from: srxmos on July 10, 2016, 01:57:10 pm

(Image removed from quote.)
Calculator allowed too

$\text{For Q13, we wish to calculate the areas of both rectangles and add them up.}\\ \text{Carefully study the diagram. The width of both rectangles is just }\frac{\pi}{6}$
$\text{The height requires us to actually find some function values.}$
$\text{When }x=\frac{\pi}{6}, \, y=\sin \frac{\pi}{6}=\frac{1}{2}\\ \text{Hence the area of the left rectangle is }A_1=\frac{1}{2}\times \frac{\pi}{6} = \frac{\pi}{12}$
$\text{When }x=\frac{\pi}{3}, \, y=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\\ \text{Hence the area of the right rectangle is }A_2=\frac{\sqrt{3}}{2}\times \frac{\pi}{6}=\frac{\sqrt{3}\pi}{12}$
$\text{Sum them up:}\\ A=\frac{\pi}{12}+\frac{\sqrt{3}\pi}{12}=\frac{(\sqrt{3}+1)\pi}{12}$

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Author Topic: Worked solution/explanation please? (Read 777 times) Tweet Share

srxmos

Worked solution/explanation please?

RuiAce

Re: Worked solution/explanation please?

RuiAce

Re: Worked solution/explanation please?

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