What is $f (x) = \int x - 3 e^{x} d x$ $f(x) = int x-3e^(x)dx$ if $f (0) = - 2$ $f(0)=-2$ ?

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Answer 1 · 2017-12-20T01:41:10

$f (x) = \frac{x^{2}}{2} - 3 e^{x} + 1$ $f(x)=x^2/2-3e^x+1$

$f (x) = \int (x - 3 e^{x}) \cdot d x = \frac{x^{2}}{2} - 3 e^{x} + C$ $f(x)=int (x-3e^x)*dx=x^2/2-3e^x+C$

After imposing $f (0) = - 2$ $f(0)=-2$ condition, I found

$C - 3 = - 2$ $C-3=-2$ , so $C = 1$ $C=1$

Thus, $f (x) = \frac{x^{2}}{2} - 3 e^{x} + 1$ $f(x)=x^2/2-3e^x+1$

What is f(x) = int x-3e^(x)dxf(x)=∫x−3exdxf(x) = int x-3e^(x)dx if f(0)=-2 f(0)=−2f(0)=-2 ?