Find the area of the region enclosed between y=2sin(x)y=2sin⁡(x) and y=3cos(x)y=3cos⁡(x) from x=0x=0 to x=0.7πx=0.7π. Hint: Notice that this region consists of two parts.

Answer 1

Let get intersection's x-coordinate: 2sinx = 3cosx; tanx = 1.5; x = tan^-11.5.

So, area A = ∫from 0 to tan^-11.5 of(3cosx - 2sinx)dx + ∫from tan^-11.5 to 0.7π of (2sinx - 3cosx)dx =

(3sinx + 2cosx) from 0 to tan^-11.5 + (- 2cosx - 3sinx) from tan^-11.5 to 0.7π ≈ ^3.9596