We have #f:RR->RR;f(a)=int_0^1abs(x-a)dx#.How to find #f'(2)#?

1 Answer
Cesareo R.
May 22, 2017

# f'(2)=1#

Explanation:

#int_0^1abs(x-a)dx = {(a < 0->int_0^1(x-a)dx=1/2-a),(0 < a < 1->int_0^a(a-x)dx+int_a^1(x-a)dx=1/2+a(a-1)),(a > 1->int_0^1(a-x)dx=a-1/2):}#

Considering now #a=2# we choose

#int_0^1abs(x-a)dx = a-1/2 rArr f'(2)=1#

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