Question

Find the area under the curves.
y=4sin(x) y=2cos(x) x=0 and .4pi
Please help!

Answer 1

Set-up an integral like \[\int\limits_{0}^{4\pi}4\sin x dx\] for each one, or use the fnInt feature on your calculator with fnInt(4*sin(x),x,0,4*pi) to spit out your answer. The second one is very similar.

Answer 2

and then add them?

Answer 3

Is it between the curves or the area under each curve?

Answer 4

Find the area of the region enclosed between

Answer 5

Ah, gotcha. Okay, so you would \[\int\limits_{0}^{.4\pi}[2\cos(x) - 4\sin (x)]dx\]. This way you subtract the larger function at x=0 by the smaller one there.

Answer 6

I tried that earlier and it didn't work! :/

Answer 7

Check your limits of integration, then. Is it really .4pi or is it 4pi? The graph crisscrosses a lot, so the integral above would only work if .4pi is the first intersection point, and according to my TI-86, it is not. Thus, I recommend splitting up the integral into two pieces: one will go from 0 to the intersection point of x=.464 as set up above, the other will go from .464 to .4pi with the order of the two functions switched.

Answer 8

It definitely is .4pi. When you mean switch you mean 4sin(x)-2cos(x)?

Answer 9

Yes, exactly, because from x=.464 to .4pi, 4sin(x) attains greater values than 2cos(x).