Q16 of 123 Page 22

The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is given by

The situation looks like this –



At the intersection points, y = 3 – x = x2 + 1


Or, x2 + x – 2 = 0


⇒ (x + 2)(x – 1) = 0


⇒ x = -2, 1 → these are our bounds


So, area A enclosed is –


A =


=


=


= 2 – 1/2 - 1/3 + 4 + 2 – 8/3


= 8 – 9/3 – 1/2


= 5 – 1/2


= 9/2 (Ans)

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