Salut, aveti o idee la problema 490...?

sinx ≤ cosx când x ∈ [0, π/4]
=> sinx ≤ sin(pi/2 - x)
=> sin[0,π/4] ≤ sin(π/4, pi/2) (A)
iar
sinx ≥ cosx cand x ∈ [π/4, π/2]
=> sinx ≥ sin(π/2-x)
=> sin[π/4, π/2] ≥ sin[0, π/4] (A)
=> sinx - cosx ≤ 0, x ∈ [0, π/4]
=> sinx - cosx ≥ 0, x ∈ [π/4, π/2]
[tex] \displaystyle \int_0^{\frac{\pi}{2}} | \sin x - \cos x| \, dx = \int_0^{\frac{\pi}{4}} (\cos x - \sin x)\, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x - \cos x)\, dx = \\ \\ = (\sin x+\cos x)\Big|_0^{\frac{\pi}{4}} + (-\cos x - \sin x)\Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \\ \\ = \sqrt{2}-1+(-1+\sqrt 2) = 2(\sqrt 2 - 1) [/tex]