The curves [tex]y=4-x[/tex] and [tex]y=x-\cos x[/tex] intersect at approximately [tex]x\approx1.8582[/tex]. The area is then
[tex]A\approx\displaystyle\int_0^{1.8582}((4-x)-(x-\cos x))\,\mathrm dx=\int_0^{1.8582}(4-2x+\cos x)\,\mathrm dx\approx4.9389[/tex]
