what is the area of the region enclosed by the graphs of y=e^x-2 y=sinx and x=0?

Answer 1

Using TI-84 we get intersection point (1.0541271, 0.86946937)

Region we have: 0 ≤ x ≤ 1.0541271; e^x - 2 ≤ y ≤ sinx

Area A = ∫₀^1.0541271(sinx - e^x + 2)dx = (- cosx - e^x + 2x)₀^1.0541271 = 2 - cos1.0541271 - e^1.0541271 + 2·1.0541271 = 0.7448