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## rosho Group Title how to find the area between y=sinx, y=cosx, y=0 between {0,pi/2} ans: 2-sqrt2 one year ago one year ago

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1. goformit100

@marcleclair

2. MarcLeclair

Cant you find the middle point? Then solve using two integrals?

3. rosho

i tried that.

4. MarcLeclair

I mean they should intersect at pi/6 ( where they have the same value) --> unit circle I believe ( I don't remember so much)

5. rosho

sqrt2/2

6. rosho

and this is calc II

7. MarcLeclair

Yeah and then integrate both int ( pi/6 - sinx ) + int ( cos x - pi/6) no? Yeah I'm taking cal II too ( well in quebec, might be different curiculum)

8. MarcLeclair

I think im totally wrong though x.x I was just trying to help. Sorry if im mistaken

9. Outkast3r09

what are you trying to do?

10. Outkast3r09

calculus or regular

11. rosho

calc 2

12. rosho

finding the area between 0, sinx and cosx on the interval of 0,pi/2

13. Outkast3r09

|dw:1363164661036:dw|

14. rosho

pretty much.

15. Outkast3r09

so you'll need to split the integral up

16. Outkast3r09

|dw:1363164746737:dw|

17. Outkast3r09

so you need the intersection point

18. rosho

already found that.

19. Outkast3r09

the reason why you need two? If you do high - low , they switch at the intersection

20. Outkast3r09

alright so what are you asking... are you done?

21. electrokid

[|dw:1363618349617:dw|

22. electrokid

$A = \int_y\int_xdxdy=\left[\int_{x=0}^{\pi\over4}\int_{y=0}^{\sin x}+\int_{x={\pi\over4}}^{\pi\over2}\int_{y=0}^{\cos x}\right]dxdy\\ A_1 = \int_{x=0}^{\pi\over4}\left[\int_{y=0}^{\sin x}dy\right]dx=\int_0^{\pi\over4}\sin x\,dx=1-{1\over\sqrt{2}}\\ A_2= \int_{x={\pi\over4}}^{\pi\over2}\left[\int_{y=0}^{\cos x}dy\right]dx=\int_{\pi\over4}^{\pi\over2}\cos x\,dx=1-{1\over\sqrt{2}}\\ A=A_1+A_2=2-{2\over\sqrt{2}}\\ A=2-\sqrt{2}$