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Find its area.y = 3 cos 5x, y = 3 − 3 cos 5x

+4 votes

 

 0 ≤ x ≤ π/5

asked Feb 3, 2013 in CALCULUS by anonymous Apprentice

2 Answers

+4 votes

Equation of the 1st curve,y1 = 3 cos 5x

Equation of the 1st curve,y2 = 3-3 cos 5x

Interval

0 ≤ x ≤ π/5

                                                                b

Recall the Area between two curves A= (y1-y2)dx

                                                               a

      π/5

A =    (3 cos 5x-(3-3 cos 5x))dx

      0

     π/5

   =    (3+6 cos 5x)dx

      0

                          π/5

    =(3x+(6/5)sin5x)

                          0

    =3(π/5)+(6/5)sin5(π/5)-0

    =(3π/5)+0

    =3π/5

Area between given two curves is 3π/5

 

Hope that helps u

 

 

answered Feb 3, 2013 by bradely Mentor
0 votes

Let, the functions are f (x) = 3 cos 5x and g (x) = 3 - 3 cos 5x , where 0 ≤ x ≤ π/5.

Because g (x) ≥ f (x) for 0 ≤ x ≤ π/5 find the area enclosed by the curves as :

Area between two curves A =  b

                                    [ g (x) - f (x) ] dx

                                  a

      π/5

A =    [ 3 - 3 cos 5x  - 3 cos 5x) ] dx

      0


   π/5

=    [ 3 - 6 cos 5x ] dx

  0

                             π/5

= [ 3x - 6/5(sin 5x) ]

                             0                     

= [ 3(π/5) - 6/5(sin 5(π/5)) ] - [ 3(0) - 6/5(sin 5(0)) ]

 

= [ 3π/5 - 6/5(sin π) ] - [ 0 - 6/5(0) ]

 

= [ 3π/5 - 6/5(0) ] - 0

 

= 3π/5 - 0

 

= 3π/5.

Area enclosed by the curves is 3π/5 square units.

answered Jun 28, 2014 by lilly Expert

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