Posted by sam on .
Find the area of the region enclosed between y=2sin(x and y=3cos(x) from x=0 to x=0.4pi
Hint: Notice that this region consists of two parts.
Notice: I'm getting 1.73762 but apparently that is wrong.

Calc 2: Area under the curve 
Mgraph,
Find x where 2sin(x)=3cos(x) (div by cos)
2tan(x)=3
tan(x)=1.5
x=56.31degr
Area(from x=0 to x=56.31)=
(3sin(56.31)+2cos(56.31)(3sin(0)+2cos(0))
=3.605552=1.60555
Area(from x=56.31 to x=72)=
(2cos(72)3sin(72))(2cos(56.31)3sin(56.31))=3.47120(3.60555)=0.13435
The total area=1.60555+0.13435=1.7399