Find the area of region enclosed by the two functions.
y= 4cos(pi x), y= 12x^2 - 3
I attempted this problem but the cos(pi x) is giving some problems.
I'm pretty sure the boundaries are [-1/2 , 1/2]
\(\displaystyle \int \limits_{-\frac 1 2}^{\frac 1 2}4\cos(\pi x)-(12x^2-3) ~dx = \\ \left . \dfrac 4 \pi \sin(\pi x)-(4x^3-3x) \right|_{-\frac 1 2}^{\frac 1 2}\)
I leave you to complete it
I'm confused on how you got the "4/pi" in front of sin(pi x)
\(\int 4 \cos(\pi x) ~dx = 4 \cdot \dfrac 1 \pi \sin(\pi x) = \dfrac{4}{\pi}\sin(\pi x)\)
+322 
