# What is #f(x) = int e^(3x)-e^(x)dx# if #f(0)=-2 #?

I found:

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To find the function ( f(x) = \int e^{3x} - e^x , dx ) given ( f(0) = -2 ), we first integrate ( e^{3x} - e^x ) with respect to ( x ) to find the antiderivative:

[ F(x) = \frac{e^{3x}}{3} - e^x + C ]

where ( C ) is the constant of integration. Then, using the given initial condition ( f(0) = -2 ), we substitute ( x = 0 ) into ( F(x) ) and equate it to ( -2 ) to find the value of ( C ).

[ F(0) = \frac{e^{3(0)}}{3} - e^0 + C = \frac{1}{3} - 1 + C = -2 ]

Solving for ( C ):

[ \frac{1}{3} - 1 + C = -2 \implies C = -2 + 1 - \frac{1}{3} = -\frac{8}{3} ]

Thus, the function ( f(x) = \int e^{3x} - e^x , dx ) with ( f(0) = -2 ) is given by:

[ f(x) = \frac{e^{3x}}{3} - e^x - \frac{8}{3} ]

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