Q. 16

# The area of the region bounded by the parabola y = x^{2} + 1 and the straight line x + y = 3 is given by

A. 45/7

B. 25/4

C. π/18

D. 9/2

Answer :

The situation looks like this –

At the intersection points, y = 3 – x = x^{2} + 1

Or, x^{2} + 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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