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find the area of the region enclosed by the curve y=cos(3x) and the interval [0,pi]?

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kmorri76 | Student, College Freshman | eNotes Newbie

Posted April 26, 2012 at 6:30 PM via web

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find the area of the region enclosed by the curve y=cos(3x) and the interval [0,pi]?

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rcmath | High School Teacher | (Level 1) Associate Educator

Posted April 28, 2012 at 3:57 PM (Answer #1)

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To find the area enclosed by the given curve we will have to graph first. We notice that between 0 and Pi/6, and Pi/2 and 5Pi/6 the graph is above the x-axis, but between Pi/6 and Pi/2, and 5pi/6 and pi the graph is below. 

To find the area we will have to use four definite integrals over the three different intervals.

`A=int_0^(pi/6)cos3xdx+int_(Pi/6)^(Pi/2)(-cos3x)dx+int_(pi/2)^(5Pi/6)cos3xdx+int_((5Pi)/6)^Pi(-cos3x)dx=`

`1/3sin3x` between o and Pi/6 + `-1/3sin3x` between Pi/6 and Pi/2+
`1/3sin3x` between Pi/2 and 5Pi/6+ `-1/3sin3x` between 5Pi/6 and Pi=

`1/3[sinPi/2-0-sin3Pi/2+sin3Pi/6+sin3(5Pi/6)-sin3pi/2-sin3pi+sin3(5Pi/6)]=`

`1/3[1-(-1)+1-(-1)-0+1]=1/3*5=5/3` 

 

 

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