Let A_1 be the bounded area enclosed by the curves y=x^2+2 ,x+y=8 and y-axis that lies in the first quadrant. Let A_2 be the bounded area enclosed by the curves y=x^2+2 ,y^2=x ,x=2, and y-axis that l…

Question

Let A_1 be the bounded area enclosed by the curves y=x^2+2 ,x+y=8 and y-axis that lies in the first quadrant. Let A_2 be the bounded area enclosed by the curves y=x^2+2 ,y^2=x ,x=2, and y-axis that lies in the first quadrant. Then A_1-A_2 is equal to

Accepted Answer

Correct answer: B. (b) For A_1 : intersection of y=x^2+2 and x+y=8 at x=2 ,y=6 A_1= _0^2 0.05em(8-x-(x^2+2)) dx= _0^2 0.05em(6-x-x^2) dx=(6x- x^22- x^33)_0^2=12-2- 83= 223For A_2 : bounded by y=x^2+2 ,y^2=x (i.e., x=y^2 ), x=2, and y-axis. In first quadrant, y^2=x goes from (0 ,0 ) to (2 , 2 ). The parabola y=x^2+2 passes through (0 ,2 ) and (2 ,6 ). A_2= _0^ 2 0.05em(2-y^2) dy=(2y- y^33)_0^ 2=2 2- 2 23= 4 23 A_1-A_2= 223- 4 23= 23 (2 2+1 )

$Let A_{1} be the bounded area enclosed by the curves y = x^{2} + 2, x + y = 8 and y -axis that lies in the first quadrant. Let A_{2} be the bounded area enclosed by the curves y = x^{2} + 2, y^{2} = x, x = 2, and y -axis that lies in the first quadrant. Then A_{1} - A_{2} is equal to$

Solution

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