What is #f(x) = int e^(2x-1)-e^(3-x)+e^x dx# if #f(2) = 3 #?

1 Answer
1s2s2p
Dec 9, 2017

#f(x)=e^(2x-1)/2+e^(3-x)+e^x+3-e^3/2+e+e^2#
#f(x)=e^(2x-1)/2+e^(3-x)+e^x+3.06456947#

Explanation:

Using #inte^(ax+b)dx=e^(ax+b)/a#

Our integration gives us:
#f(x)=e^(2x-1)/2+e^(3-x)+e^x+C#

We also know that #e^(2(2)-1)/2+e^(3-2)+e^2+C=3#

#e^3/2+e+e^2+C=3#

#C=3-e^3/2+e+e^2~~3.06456947#

#f(x)=e^(2x-1)/2+e^(3-x)+e^x+3-e^3/2+e+e^2#
#f(x)=e^(2x-1)/2+e^(3-x)+e^x+3.06456947#

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