sketch the region enclosed by the given curves and find its area.
y= cos(pix) y=4x^(2)-1. you dont have to sketch the graph out just explain to how to work it out

Question

sketch the region enclosed by the given curves and find its area.   
y= cos(pix) y=4x^(2)-1. you dont have to sketch the graph out just explain to how to work it out

el_arrow · Answer

$$y=\cos \pi x  $$

el_arrow · Answer

and y=4x^(2)-1

el_arrow · Answer

was up man could you help me out please

anonymous · Answer

I don't think there's anyway to know right off the bat where the curves intersect, but WA tells us they intersect at $x=\pm\dfrac{1}{2}$ http://www.wolframalpha.com/input/?i=Cos%5BPi+x%5D%3D4x%5E2-1

el_arrow · Answer

sorry i had to refresh somethings going on with my connection

el_arrow · Answer

so the only way to know for sure is by graphing it

anonymous · Answer

Not necessarily. I just didn't see a quick algebraic method to finding the roots to that equation. As for checking what the actual bounded area is (in general), plotting is probably the easiest way.

So if you managed to get that far on your own, great! Let's assume you had found the intersections without plotting. Now you need to know which curve is the "greater/upper" curve, and which is the "lower" one.

Without plotting, you know that the bounded region referred to in the question falls in one of three possible intervals: $\left(-\infty,-\dfrac{1}{2}\right)$, $\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$, and $\left(\dfrac{1}{2},\infty\right)$. If you know about the functions' behaviors, you'll find that the second interval is the only plausible one, since it is finite.

el_arrow · Answer

how do i know which is upper or lower?

anonymous · Answer

So now you have to check which curve is higher than the other. Pick a convenient value from the interval (-1/2, 1/2), like $x=0$, and evaluate each function at that point.
$$\cos\pi(0)=\cos0=1\
4(0)^2-1=0-1=-1$$
which means $\cos\pi x$ is above $4x^2-1$.

This means the area is given by the integral,
$$A=\int_{-1/2}^{1/2}\left(\cos\pi x-\left(4x^2-1\right)\right)~dx$$

el_arrow · Answer

okay so you just plug in any value

anonymous · Answer

Any value in the interval, yes. Not the endpoints, and not some value outside of it.

el_arrow · Answer

and the greatest number is the upper and smallest is lower

anonymous · Answer

Yep

el_arrow · Answer

okay i got it thanks man

anonymous · Answer

yw