求曲线y=sinx和y=cosx在[0,3π／2]上围成的区域的面积
1个回答
在[0,π/4]上,cosx > sinx
在[π/4,5π/4]上,sinx > cosx
在[5π/4,3π/2]上,cosx > sinx
在[0,3π/2]上围成的区域的面积
S = S1 [0,π/4]
+ S2 [π/4,5π/4]
+ S3 [5π/4,3π/2]
= ∫(cosx - sinx)dx [0,π/4]
+ ∫(sinx - cosx)dx [π/4,5π/4]
+ ∫(cosx - sinx)dx [5π/4,3π/2]
= (sinx + cosx) [0,π/4]
+ (-cosx - sinx) [π/4,5π/4]
+ (sinx + cosx) [5π/4,3π/2]
= sin(π/4) + cos(π/4) - sin0 - cos0
- [cos(5π/4) + sin(5π/4)] + [cos(π/4) + sin(π/4)]
+ sin(3π/2) + cos(3π/2) - sin(5π/4) - cos(5π/4)
= 1/√2 + 1/√2 - 0 - 1
- (-1/√2 - 1/√2) + (1/√2 + 1/√2)
+ (-1 + 0) - (-1/√2 - 1/√2)
= √2 - 1 + √2 + √2 + √2 - 1
= 4√2 - 2
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