The area enclosed by

Question

The area enclosed by y = x² – x and y = 1 – x² is

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Answer 1

1-x^2 - (x^2 -x)

= 1-2x^2 +x

take the integral

= -2x^3/3 +x^2/2 + x

evaluate from where the 2 parabolas intersect x =-1/2 and 1

set them equal

1-x^2 = x^2 -x

combine like terms

2x^2 -x -1 = 0

factor

(2x+1)(x-1) =0

set eqch factor equal to 0

2x+1=0, x-1=0

solve for x

x =1 and x =-1/2

substitute those values for x in the integral and subtract

-2(1)/3 + 1/2 +1 - (-2/24 + 1/8 -1/2)

= -2/3 +3/2 - (2/24 + 3/24 -12/24)

= 5/6 - (-7/24)

= 20/24 +7/24

=27/24

= 9/8

Answer 2

The graphs intersect at x = - 1/2 and x = 1 , which are the bounds of integration.

On this interval, the graph of y = 1 - x² is above the other graph.

A = ∫_-1/2¹ [(1 - x²) - (x² - x)] dx

= ∫_-1/2¹ [1 - 2x² + x] dx

= x - 2/3x³ + 1/2x² ]_-1/2¹

= 5/6 - ( - 7/24) = 27/24 = 9/8