## sauravshakya 2 years ago y=sinx find the average value of y in the interval 0 to 90

1. sauravshakya

I mean find lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(n+1)} where ndx=90

2. sirm3d

mean value theorem for integrals?

3. sauravshakya

No not mean value theorem... Here I just want to find the avereage value

4. sauravshakya

ndx=90 n=90/dx Now, lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(n+1)} lim x-->0 {sin0 +sindx +sin2dx +sin3dx +...+sin(ndx)/(90/dx +1)} lim x-->0 {(sin0 +sindx +sin2dx +sin3dx +...+sin(ndx))dx/(90+dx)} |dw:1359042392858:dw|

5. sauravshakya

Now, from there I will get 1/90 Which surely not CORRECT

6. sauravshakya

Now, If I do ndx =pi/2 then I will get 2/pi

7. sauravshakya

@experimentX can u PLZ help

8. experimentX

this should be close to zero.|dw:1359042684509:dw|

9. shubhamsrg

Well 90 is degree, you surely can't use that in these calculations ? pi/2 is appropiate.

10. experimentX

hmm ... is 90 really degree?

11. sauravshakya

And yes that was what I was going to ask... Why does it matter

12. experimentX

degree is just a notation made for the sake of ease ... radian is the actual value.

13. experimentX

you might have noted lim x->0 sin(x)/x = 1 or ... as approx .. when 'x' is very small, sin(x) ~ x ... this never works for degree ... it sounds funny.

14. sauravshakya

15. sauravshakya

And we supposed pi/2 radian =90 degree

16. experimentX

yes!! radian is more natural measure of angles.

17. sauravshakya

So, is the correct answer 2/pi

18. experimentX

yep!!

19. experimentX

compare these with the area bound by the curves with x axis an 0 and pi/2

20. sauravshakya

thanx again

21. experimentX

yw