What is $f (x) = \int (x^{2} - 2 x) (e^{x} - 3 x) d x$ $f(x) = int (x^2-2x)(e^x-3x) dx$ if $f (1) = 2$ $f(1 ) = 2$ ?

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Answer 1 · 2017-07-24T18:28:45

$f (x) = (x^{2} - 4 x + 4) \cdot e^{x} - (\frac{3}{4} \cdot x^{4} - 2 x^{3}) + \frac{3}{4} - e$ $f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e$

$f (x) = \int (x^{2} - 2 x) \cdot e^{x} \cdot d x - \int (3 x^{3} - 6 x^{2}) \cdot d x$ $f(x)=int(x^2-2x)*e^x*dx-int(3x^3-6x^2)*dx$

= $[(x^{2} - 2 x) - (2 x - 2) + 2] \cdot e^{x} - (\frac{3}{4} \cdot x^{4} - 2 x^{3}) + C$ $[(x^2-2x)-(2x-2)+2]*e^x-(3/4*x^4-2x^3)+C$

= $(x^{2} - 4 x + 4) \cdot e^{x} - (\frac{3}{4} \cdot x^{4} - 2 x^{3}) + C$ $(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+C$

After imposing $f (1) = 2$ $f(1)=2$ condition,

$(1 - 4 + 4) \cdot e - (\frac{3}{4} - 2) + C = 2$ $(1-4+4)*e-(3/4-2)+C=2$

$C = \frac{3}{4} - e$ $C=3/4-e$

Thus, $f (x) = (x^{2} - 4 x + 4) \cdot e^{x} - (\frac{3}{4} \cdot x^{4} - 2 x^{3}) + \frac{3}{4} - e$ $f(x)=(x^2-4x+4)*e^x-(3/4*x^4-2x^3)+3/4-e$

1) I integrated the right side of the equation.

2) I found C for $f (1) = 2$ $f(1)=2$ condition.

What is f(x) = int (x^2-2x)(e^x-3x) dxf(x)=∫(x2−2x)(ex−3x)dxf(x) = int (x^2-2x)(e^x-3x) dx if f(1 ) = 2 f(1)=2f(1 ) = 2 ?