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

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Answer 1 · 2017-02-20T22:18:39

The function #f# is:

#f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})#.

By integrating the value of #f#:

#f (x) = int (e^{2 x - 1} - e^{1 - 2x} + e^x) dx =#

#= int e^{2x-1} dx - int e^{1-2x} dx + int e^x dx =#

#= 1/2 e^{2x-1} - (- 1/2) e^{1 - 2x} + e^x + C =#

#= 1/2 (e^{2x-1} + e^{1-2x}) + e^x + C#

Then, substituting the known point of #f#:

#f (2) = 1/2 (e^3 + e^{- 3}) + e^2 + C = 3#,

we can obtain the value of the constant #C#:

#C = 3 - e^2 - 1/2 (e^3 + e^{- 3})#.

The function #f# will have the following formula:

#f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})#.