f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
来源：学生作业帮助网 编辑：六六作业网 时间：2025/01/30 07:52:51
f''(x)=sin(x-1)^2f(0)=0求∫(0,1)f(x)dxf''(x)=sin(x-1)^2f(0)=0求∫(0,1)f(x)dxf''(x)=sin(x-1)^2f(0)=0求∫(0,1)f
f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
做个简单点的吧,设f'(x)=[sin(x-1)]^2
f(x)=∫f'(x)dx
=∫[sin(x-1)]^2dx
=∫[sin(x-1)]^2d(x-1)
=∫{1-cos[2(x-1)]}/2*d(x-1)
=(x-1)/2-1/4*sin[2(x-1)]+C
f(0)=-1/2-1/4*sin(-2)+C=0
=> C=1/2-1/4*sin2
f(x)=x/2-1/4*sin[2(x-1)]-1/4*sin2
∫ (0,1) f(x)dx
=∫ (0,1) {x/2-1/4*sin[2(x-1)]-1/4*sin2}dx
=(0,1) {x^2/4+1/8*cos[2(x-1)]-x/4*sin2}
={1/4+1/8-1/4*sin2}-{0+1/8*cos2-0}
=(3-2sin2-cos2)/8
如果是sin[(x-1)^2]就没法做
设F(x)>0,F(0)=1,F‘(x)=f(x),且f(x)F(x)=sin^2(2x)(x>=0),求f(x).是F'（x)=f(x)
已知函数f(x)=1-sin πx/2,则f(0)+f(1)+f(2)+.+f(2010)=
f(sin 2/x )=1+cosx 求f(x)
若f(x)=sin πx/6,则f(1)+f(2)+...+f(102)=?
若f(x)=sin(π/4)x,求f(1)+f(2)+.+f(2010)
f(x)为奇函数,x>0,f(x)=sin 2x+cos x,则x
f(x)={(1-e^x)/sin(x/2) ,x>0 ;ae^2x,x
复合函数f（x）=x^2sin(1/x) （x>0） f(x)=0 x
若f(x)=sinπ/2x,则f(1)+f(2)+f(3)+...+f(2012)+f(2013)=
已知函数f(x)=-1/2+sin(5/2x)/2sin(x/2)(0
f'(cos^2x)=sin^2x ,f(0)=0,求f(x)
f'(sin x)=1-cos x 求f''(x)
f(x)=sin-1次方 x ,求f'(x)
f(sin^2x)=x/sinx 求f(x)
已知f(sin-1)=cos2x+2,求f(x)
f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
若f'(cos²x)=1+2sin²x,且f(0)=1,求f(x)
证明∫( 0,π／2 ) (f sin x/(f sin x+f cos x) dx=π /4