What is $f (x) = \int 5 x - 2 x e^{x} d x$ $f(x) = int 5x-2xe^(x)dx$ if $f (0) = - 2$ $f(0)=-2$ ?

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Answer 1 · 2017-07-11T17:42:46

$f (x) = \int 5 x \cdot d x - \int 2 x \cdot e^{x} \cdot d x$ $f(x)=int5x*dx-int2x*e^x*dx$

$= \frac{5}{2} \cdot x^{2} - (2 x \cdot e^{x} - \int 2 e^{x} \cdot d x)$ $=5/2*x^2-(2x*e^x-int2e^x*dx)$

$= \frac{5}{2} \cdot x^{2} - (2 x - 2) \cdot e^{x} + C$ $=5/2*x^2-(2x-2)*e^x+C$

Let $x = 0$ $x=0$ , $C + 2 = - 2$ $C+2=-2$ , hence $C = - 4$ $C=-4$ .

Thus, $f (x) = \frac{5}{2} \cdot x^{2} - (2 x - 2) \cdot e^{x} - 4$ $f(x)=5/2*x^2-(2x-2)*e^x-4$

1) Take integral.

2) Impose $f (0) = - 2$ $f(0)=-2$ condition.

What is f(x) = int 5x-2xe^(x)dxf(x)=∫5x−2xexdxf(x) = int 5x-2xe^(x)dx if f(0)=-2 f(0)=−2f(0)=-2 ?