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Let f(x)= cases 13, & x 2\ b(1- x)( -2x)2, & x> 2 cases. If f is continuous at x= 2, then the value of 03b-6 |x2+2x-3 |dx is :
(4) 4
Put x= 2+h, h 0+: 1- x=1- h, ( -2x)2=4h2
h 0+ b(1- h)4h2= b4 12= b8
Continuity at 2: b8= 13 b= 83 3b-6=2
x2+2x-3=(x+3)(x-1): negative on (0, 1), positive on (1, 2)
[figure]
I= 01 (3-2x-x2 )dx+ 12 (x2+2x-3 )dx
= [3x-x2- x33 ]01+ [ x33+x2-3x ]12= 53+ ( 23+ 53 )=4
Answer: (4)
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