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,）

∫[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) = 1a+π/4 = π/2a = π/4时,f(a)取得最大值并且最大值为f(π/4) = √2*sin(...∫[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！cosx的积分是sinxsinx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