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Автор:  jagdish [ 08 фев 2012, 17:32 ]
Заголовок сообщения:  area

If [math]A[/math] be the area bounded by the curve [math]f(x) = \left|\frac{\sin x+\cos x}{x}\right|[/math] and X- axis

and line [math]x=\pi[/math] and [math]x=3\pi[/math]. Then Max. and Min. value of [math]A[/math] is

Автор:  igor_vis [ 10 фев 2012, 12:24 ]
Заголовок сообщения:  Re: area

Min. value of A is 0

f(x)=(sin(x)+cos(x))/x (if (sin(x)+cos(x))/x > 0)
f(x)=-(sin(x)+cos(x))/x (if (sin(x)+cos(x))/x < 0)


max(f(x)) <= sqrt(2)/pi = 0,450158158 - first approximation
max(f(x)) = 0,372706795 (by Ms Excel)

Автор:  jagdish [ 10 фев 2012, 19:20 ]
Заголовок сообщения:  Re: area

would you like to explain it to me

Thanks

Автор:  igor_vis [ 12 фев 2012, 08:26 ]
Заголовок сообщения:  Re: area

(sin(x)+cos(x)) =(sin(x)+sin(pi - x)) = 2*sin((x+(pi/2 - x))/2)*cos((x-(pi/2 - x))/2) =
=2*sin(pi/4)*cos(x-pi/4) = 2*sqrt(2)/2*cos(x-pi/4) = sqrt(2)*cos(x-pi/4) <= sqrt(2)
(sin(x)+cos(x)) / x <= sqrt(2) / min(x) = sqrt(2) / pi - first approximation
for the exact solution of the problem we have to find the derivative of the function and equate it to zero
f(x)=|g(x)|
min(f(x))=0 if g(x) = 0 if cos(x-pi/4)= 0 (for example x-pi/4 = 3pi/2 => x = 7pi/8 => f(x)=g(x)=0
max(f(x)) -???
g(x)=(sin(x)+cos(x)) / x = sqrt(2)*cos(x-pi/4) / x
g`(x)= sqrt(2)*(-sin(x-pi/4)*x-cos(x-pi/4))/ x^2
g`(x)=0 if ctg(x-pi/4)=-x
to get the exact answer, we must find solutions
ctg(x-pi/4)=-x
I can not solve this equation, I'm sorry

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