since 0.7π = 2.199, and
(a) 3cosx > 2sinx for 0 <= x <= 0.983
(b) 2sinx > 3cosx for 0.98279 <= x <= 2.199
The area is
∫[0,0.983] (3cosx-2sinx) dx + ∫[0.983,2.199] (2sinx-3cosx) dx
Find the area of the region enclosed between y=2sin(x) and y=3cos(x) from x=0 to x=0.7π. Hint: Notice that this region consists of two parts.
1 answer
1 answer