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To find the points of intersection of the curves, we solve the equations of the curves simultaneously. Therefore x^2 - 4 = 2x - x^2 x^2 - 4 - 2x + x^2 = 0 2x^2 - 2x - 4 = 0 Divide by 2 x^2 - x - 2 = 0 (x+1)(x-2) = 0 x = -1, 2

The area A = |integral (x^2 - 4) - (2x - x^2) dx from -1 to 2| = |integral x^2 - 4 - 2x + x^2 dx from -1 to 2| = |integral 2x^2 - 2x - 4 dx from -1 to 2| = | 2x^3/3 - 2x^2/2 - 4x from -1 to 2 | = | 2x^3/3 - x^2 - 4x from -1 to 2 | = | (2 * 2^3/3 - 2^2 - 4*2) - [2*(-1)^3/3 - (-1)^2 - 4(-1)] | = |(16/3 - 4 - 8) - (-2/3 - 1 + 4)| = |16/3 - 12 + 2/3 + 1 - 4| = |16/3 + 2/3 - 12 + 1 - 4| = |18/3 - 15| = |6 - 15| = |-9| = 9 Ans: 9 Note: | | stands for absolute value.

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