What is #f(x) = int 1/x-e^x dx# if #f(1) = 0 #?

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Answer 1 · 2016-05-05T03:59:25

#f(x)=ln(x)-e^x+e#

#f(x)= int(1/x-e^x)dx#
#=int1/xdx-inte^xdx# (Linearity)

#int1/xdx = ln(x)# (Standard integral)

#inte^x = e^x# (Exponential rule)

Therefore:
#f(x) = ln(x)-e^x +C# (Where C is the constant of integration)

We are told that #f(1)=0# Thus:
#ln(1)-e^1+C=0#

#0 - e +C =0#

#C=e#

Hence: #f(x)=ln(x)-e^x+e#