题目
设y=∫(0到x)(sint)^(1/2)dt(0<=x<=pai),则曲线y的长度是
提问时间:2020-10-30
答案
y'=(sinx)^(1/2)
ds=[1+sinx]^(1/2)dx
s=∫(0,π)[1+sinx]^(1/2)dx=∫(0,π)[sin(x/2)+cos(x/2)]dx=2[sin(x/2)-cos(x/2)](0,π)=4
ds=[1+sinx]^(1/2)dx
s=∫(0,π)[1+sinx]^(1/2)dx=∫(0,π)[sin(x/2)+cos(x/2)]dx=2[sin(x/2)-cos(x/2)](0,π)=4
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