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

2. Originally Posted by waite3
Find the area of the region enclosed between y=4sin(x) and y=3cos(x) from x=0 to x=0.9pi .
the area will be given by $\int_0^{0.9 \pi} (\mbox{Top function } - \mbox{ Bottom function})~dx$

so which is which?

note that the functions switch roles at some point in the interval

3. the sin function is the top and we get 6.877, yet the online program for submission being used is saying that that answer is incorrect, whereabouts would my error be lying? thanks for the help

4. why dont you graph the two functions? you will see why your answer is wrong.

hint: $cos(0)=1$ while $sin(0)=0$