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

Question

 

 0 ≤ x ≤ π/5

Answer 1

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

 

 

Answer 2

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.