Question

已知a>0,当∫（cosx-sinx)dx（上限a，下限O）取最大值时，a的最小值为___________(要详细过程，谢谢！）

Answer 1

∫[0->a) (cosx-sinx) dx
= cosx+sinx(0->a)
= cosa+sina-cos0-sin0
= cosa+sina-1

设f(a) = cosa+sina-1
= √2*sin(a+π/4) - 1
当sin(a+π/4) = 1
a+π/4 = π/2
a = π/4时，f(a)取得最大值
并且最大值为f(π/4) = √2*sin(π/4+π/4) - 1
= √2 - 1

∴a的值为π/4

追问

∫[0->a) (cosx-sinx) dx
= cosx+sinx(0->a)
= cosa+sina-cos0-sin0
= cosa+sina-1

这个过程我不大理解，请解释一下，谢谢！

追答

cosx的积分是sinx
sinx的积分是-cosx
所以积分∫(cosx-sinx) = ∫cosx dx - ∫sinx dx
= sinx - (-cosx)
= sinx + cosx
求得原函数后再代入上限和下限,即∫(a->b) f(x) dx = F(b) - F(a)，F(x)是f(x)的原函数
= (sinx + cosx) |(上限a，下限0)
= (sina + cosa) - (sin0 + cos0)，上限a代入的原函数再减去下限0代入的原函数
= (sina + cosa) - (0 + 1)
= sina + cosa - 1