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

Answer 1

Elijah Abram

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

First we have to integrate

#int(e^(2x-1)-e^(3-x)+e^(-x))dx#

the integral is

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

but #f(2)=3#

so, it follows

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

Solving for C integration constant

#C=-e^3/2-e+e^(-2)+3#

Our final answer

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

God bless....I hope the explanation is useful.

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Answer 2

Charles Arocha

The function ( f(x) = \int e^{2x-1} - e^{3-x} + e^{-x} , dx ) satisfies ( f(2) = 3 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.