作业帮设f(x)=∫＜x,x+π/2＞|sint|dt,求f(x)在区间[-41π/4,41π/4]上的最大值与最小值
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![设f(x)=∫＜x,x+π/2＞|sint|dt,求f(x)在区间[-41π/4,41π/4]上的最大值与最小值](/uploads/image/z/15248204-44-4.jpg?t=%E8%AE%BEf%28x%29%3D%E2%88%AB%EF%BC%9Cx%2Cx%2B%CF%80%2F2%EF%BC%9E%7Csint%7Cdt%2C%E6%B1%82f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B-41%CF%80%2F4%2C41%CF%80%2F4%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%8E%E6%9C%80%E5%B0%8F%E5%80%BC)
设f(x)=∫＜x,x+π/2＞|sint|dt,求f(x)在区间[-41π/4,41π/4]上的最大值与最小值
设f(x)=∫＜x,x+π/2＞|sint|dt,求f(x)在区间[-41π/4,41π/4]上的最大值与最小值
设f(x)=∫＜x,x+π/2＞|sint|dt,求f(x)在区间[-41π/4,41π/4]上的最大值与最小值
sint的对称轴为x=π/2+kπ,k∈Z,对称中心为(kπ,0),k∈Z.
x+π/2-x=π/2,因而f(x)在R上的最大值与最小值分别为
f(π/2+kπ-(π/2)/2)=f(kπ+π/4),
f(kπ-(π/2)/2)=f(kπ-π/4).
令kπ+π/4∈[-41π/4,41π/4],
则解得k∈[-21/2,10],
令kπ-π/4∈[-41π/4,41π/4],
则解得k∈[-10,21/2],
所以k∈[-10,10],k∈Z.
因而f(x)在区间[-41π/4,41π/4]上的最大值与最小值就是f(x)在R上的最大值与最小值
maxf(x)=f(kπ+π/4)=∫＜kπ+π/4,kπ+π/4+π/2＞|sint|dt=∫＜π/4,π/4+π/2＞|sint|dt
=∫＜π/4,3π/4＞sintdt=-cos(3π/4)-(-cos(π/4))=√2/2+√2/2=√2
minf(x)=)=∫＜kπ-π/4,kπ-π/4+π/2＞|sint|dt=∫＜-π/4,-π/4+π/2＞|sint|dt
=∫＜-π/4,π/4＞|sint|dt=∫＜-π/4,0＞|sint|dt+∫＜0,π/4＞|sint|dt=2∫＜0,π/4＞sintdt
=2(-cos(π/4)-(-cos0))=2(-√2/2+1)=2-√2
f(x)在区间[-41π/4,41π/4]上的最大值为√2,最小值2-√2.