Question

求大神解答44题，需要详细的解题步骤与解析，在线的，谢谢！！！

Answer 1

画一下图像，看出是对称的，计算不是很难

Answer 2

∫(0->π/2)|sinx-cosx|dx
=∫(0->π/4)(cosx-sinx)dx+∫(π/4->π/2)(sinx-cosx)dx
=sinx+cosx|(0->π/4)+(-cosx-sinx)|(π/4->π/2)
=√2-1+√2-1
=2(√2-1)

Answer 3

因为在0到π/4的区间sinx-cosx的值是负数，而在π/4到π/2sinx-cosx的值是正数，
所以∫(0→π/2) |sinx-cosx|dx
=∫(0→π/4) (cosx-sinx)dx + ∫(π/4→π/2) (sinx-cosx)dx
=(sinx+cosx) | (0→π/4) + (-cosx-sinx) | (π/4→π/2)
=(√2 -1) + (-1+√2)=2(√2 -1)