The correct option is D 8
pointofintersectionofgivencurves
(x−3)2=x
⇒x2+9−6x−x=0
⇒x2−7x+9=0
x=7±√49−362
=5.303,1.697
RequiredArea=b∫a(uppercurve)−(lowercurve)dx
A=5.303∫1.697[(x)−(x−3)2]dx
=5.303∫1.697[(x)−(x2+9−6x)]dx
=5.303∫1.697[−x2−9+7x]dx
=[−x33+7x22−9x]5.3031.697
=[(−(5.303)33+7(5.303)22−9(5.303))−(−(1.697)33+7(1.697)22−9(1.697))]
=7.811