Vei trasa graficul celor 2 puncte,Stabilesti punctul de intersectie rezolvand ecuatia sin x=cosx x=π/4
Vom deosebi trei cazuri x∈[0,π/4] g(x)>f(x) vom calcula A1
A1=∫cos x-sinx)*dx=∫cosdx-∫sinxdx=sin x+cos x x∈[0,π/4]=conf Leibnitz Newton =sinπ/4+cosπ/4-sino-cos0=√2-1
Pe intervalul [π/4 .π/2] sinx >cos x
A2=∫(sinx-cosx)*dx=∫sinxdx-∫cosxdx=-cosx-sin x=
-cosπ/2-sinπ/2+cosπ/4+sin π/4=-1+√2
Obsrvi ca pt x∈[π/2 , π] gx<0. De aceea vom pune - in fata integralei A3
A3=-∫cosxdx=-sinx x∈[π/2, π]
A3=-sinπ+sin π/2=1
A=A1+A2+A3=1
