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f'x=sin(x-1)^2 f0=0 求∫ (0,1) f(x)dx
人气:400 ℃ 时间:2020-07-03 23:32:49
解答
f(x)=∫ (0,x) sin(y-1)^2dy∫ (0,1) f(x)dx=∫ (0,1)dx∫ (0,x) sin(y-1)^2dy=∫ (0,1)dy∫ (y,1) sin(y-1)^2dx=∫ (0,1)(1-y)sin(y-1)^2dy=(1/2)cos(y-1)^2|(0,1)=(1/2)(1-cos1)第三个等号开始不太懂=∫ (0,1)(1-y)sin(y-1)^2dy=-(1/2)∫ (0,1)sin(y-1)^2d(y-1)^2(相当于sinxdx)=(1/2)cos(y-1)^2|(0,1)=(1/2)(1-cos1)
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