We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
41
3
avatar+322 

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+5097 
+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+5097 
+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+5097 
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

5 Online Users