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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] 

 May 1, 2019

Best Answer 

 #1
avatar+6196 
+1

\(\displaystyle \int \limits_{-\frac 1 2}^{\frac 1 2}4\cos(\pi x)-(12x^2-3) ~dx = \\ \left . \dfrac 4 \pi \sin(\pi x)-(4x^3-3x) \right|_{-\frac 1 2}^{\frac 1 2}\)

 

I leave you to complete it

 May 1, 2019
 #1
avatar+6196 
+1
Best Answer

\(\displaystyle \int \limits_{-\frac 1 2}^{\frac 1 2}4\cos(\pi x)-(12x^2-3) ~dx = \\ \left . \dfrac 4 \pi \sin(\pi x)-(4x^3-3x) \right|_{-\frac 1 2}^{\frac 1 2}\)

 

I leave you to complete it

Rom May 1, 2019
 #2
avatar+322 
0

I'm confused on how you got the "4/pi" in front of sin(pi x) 

Ruublrr  May 1, 2019
 #3
avatar+6196 
0

\(\int 4 \cos(\pi x) ~dx = 4 \cdot \dfrac 1 \pi \sin(\pi x) = \dfrac{4}{\pi}\sin(\pi x)\)

Rom  May 2, 2019

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