Find the area of the region enclosed between y=4sin(x) and y=3cos(x) from x=0 to x=0.9p. Hint: Notice that this region consists of two parts?

Question

how do I solve this?

Answer 1

4Sinx and 3Cosx across each other at

4Sinx = 3Cosx

Tanx = 3/4  implies x = 0.2pi

 

From 0 to 0.2pi, 3Cosx > 4 Sinx

Area 1 = Fixed integration of (3Cosx-4Sinx), from 0 to 0.2pi

 

From 0.2pi to 0.9pi, 4Sinx > 3Cosx

Area 2 = Fixed integration of (4Sinx-3Cosx), from 0.2pi to 0.9pi

 

Total Area = Area 1 + Area 2