作业帮f'(x)=sin(x-1)^2 f(0)=0 求∫ (0,1) f(x)dx
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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
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