region enclosed

[CellCoree]

Find the area of the region enclosed between y=3sin(x) and y=4cos(x) from x=0 to x=0.9\pi .


does that mean the integral would like this: \int 4cos(x) - 3sin(x) (a=0, b = 0.9*pi) is that correct?

[Galileo]

The cosine is larger than the sine at x=0 and the sine is larger than the cosine at 0.9\pi so they cross each other somewhere in that interval.
Since area is always positive you should split up your integral, and find at which point they intersect.
A picture may help.

[CellCoree]

The cosine is larger than the sine at x=0 and the sine is larger than the cosine at 0.9\pi so they cross each other somewhere in that interval.
Since area is always positive you should split up your integral, and find at which point they intersect.
A picture may help.


ah, 0.9pi is equal to 162 degrees right? i graphed it using my calculator. dont know why, but i thought of 0.9pi as 2.872433.

so would my equation look like this...

\int 4cos(x) -3sin(x) (a=0,b=.295\pi) + \int 3sin(x) - 4cos(x) (a=.295\pi,b=9\pi) is that correct?

[Galileo]

The notation is a bit weird, but I think you got the right idea.
\int_0^{0.295\pi}(4\cos x-3\sin x)dx+\int_{0.295\pi}^{0.9\pi}(3\sin x-4\cos x)dx

[CellCoree]

lol yeah, i didnt know how to insert a and b into the integral sign. thanks for the help