We have #f:RR->RR# a differentiable function with #f(1)=0# and #f'(e^x)=x-2,forall x>=0#.How to calculate #int_1^2f(x)dx#?

1 Answer
Cesareo R.
May 30, 2017

# 2 log_e2-23/12#

Explanation:

If #g(e^x)=x-2 rArr g(e^(log_e x))=log_e x-2# then

#g(x) = log_e x-2 = f'(x)# then

#f(x) =xlog_ex-x-x^2/2+C#

but

#f(1)=0 rArr -1-1/2+C=0 rArr C = 3/2# and finally

#int_1^2 f(x)dx = int_1^2 (xlog_ex-x-x^2/2+3/2)dx = 2 log_e2-23/12#

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