If the area bounded by y = x 2 and y = 2 1 + x 2 is K 1 π - K 2 3 sq. units (where K 1 , K 2 ∈ Z ), then the value of K 1 + K 2 is equal to

Step-by-step solution

Drawing graphs of y = x 2 and y = 2 x 2 - 1 For points of intersection, x 2 = 2 x 2 + 1 ⇒ x 4 + x 2 - 2 = 0 x 2 = 1 ⇒ x = ± 1 Required area = 2 ∫ 0 1 2 1 + x 2 - x 2 d x = π - 2 3 s q . u n i t s K 1 = 1 ,   K 2 = 2 ⇒ K 1 + K 2 = 3