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

1 Answer
benjidunn
Mar 30, 2016

#f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+3.768#

Explanation:

First integrate to obtain #f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+c#
Then substitute #f(2)=3# to obtain #3=1/2e^3-1/3e^4+e^2+c#
Rearrange for c then evaluate: #c=3-1/2e^3+1/3e^4-e^2=3.768# (4s.f.)
Thus the original function is #f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+3.768#

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