Solving the equations of the curves to get: sinx=cosx →sinx/cosx =tanx=1 ,cosx≠0 ∴x=π/4 or x=5π/4 Thus, the required area is: A=∫_0^(π⁄4)〖(cosx-sinx)〗 dx+∫_(π⁄4)^(5π⁄4)〖(sinx-cosx)〗 dx+∫_(5π⁄4)^2π〖(cosx-sinx)〗 dx =[sinx+cosx ]_0^(π⁄4)+[〖-cos〗x-sinx ]_(π⁄4)^(5π⁄4)+[sinx+cosx ]_(5π⁄4)^2π =(sin〖π/4〗+cos〖π/4〗 )-1+(-cos〖5π/4〗-sin〖5π/4〗 )-(-cos〖π/4〗-sin〖π/4〗 )+(sin2π+cos2π )-(sin〖5π/4〗+cos〖5π/4〗 ) =1/√2+1/√2-1+(1/√2+1/√2)+(1/√2+1/√2)+1+1/√2+1/√2=8/√2=4√2 square units

Solving the equations of the curves to get:

need an explanation for this answer? contact us directly to get an explanation for this answersinx=cosx →sinx/cosx =tanx=1 ,cosx≠0

∴x=π/4 or x=5π/4

Thus, the required area is:

A=∫_0^(π⁄4)〖(cosx-sinx)〗 dx+∫_(π⁄4)^(5π⁄4)〖(sinx-cosx)〗 dx+∫_(5π⁄4)^2π〖(cosx-sinx)〗 dx

=[sinx+cosx ]_0^(π⁄4)+[〖-cos〗x-sinx ]_(π⁄4)^(5π⁄4)+[sinx+cosx ]_(5π⁄4)^2π

=(sin〖π/4〗+cos〖π/4〗 )-1+(-cos〖5π/4〗-sin〖5π/4〗 )-(-cos〖π/4〗-sin〖π/4〗 )+(sin2π+cos2π )-(sin〖5π/4〗+cos〖5π/4〗 )

=1/√2+1/√2-1+(1/√2+1/√2)+(1/√2+1/√2)+1+1/√2+1/√2=8/√2=4√2 square units