help please with calculus problem

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determine the area enclosed by the curves f(x)=x^2-2x and g(x)=4-x^2

Jul 22nd, 2015

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f(x)=x^2-2x and g(x)=4-x^2

x^2-2x =  4-x^2

2x^2 - 2x -4 = 0

x^2 - x - 2 =0

both curves intersect at x = -1 and x = 2


so area of two curves is ( lower limit =-1 and upper limit = 2)

      2

    ∫    (4-x^2) - (x^2-2x)                            

-1

                                                                     2

=   [ ( 4x - x ^3 /3 )  -  ((x^3/ 3 ) - 2 x^2/2))  ]

                                                                     -1


the rest will be in the next submit.....please wait 5 minutes

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 22nd, 2015

                                                               2

= [  4x - x ^3 /3 )  -  (x^3/ 3 ) + 2 x^2/2 ]

                                                              -1

=  [4 *2 - 2^3 /3 - 2^3 /3 +  2 * 2^2/2 ]   -   [ 4 * -1 - ( -1 ) ^3 / 3 - ( (-1)^3 /3 ) + 2* (-1) ^2 /2 ] 

= [ 8- 8/ 3- 8/3  + 4  ]  -  [ -4  + 1/3  + 1/3 + 1]

 =  8 +4 +4 -6 -1

 = 9.................................................................ANSWER


area between  f(x)=x^2-2x and g(x)=4-x^2 =   9 ..............................................answer

Jul 22nd, 2015

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