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

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Answer 1 · 2018-08-12T00:46:20

$I = \int (x e^{2 - x} - 2 x^{2} + 2) d x$ $I=int(xe^(2-x)-2x^2+2)dx$

$= \int x e^{2 - x} d x - 2 \int x^{2} d x + 2 \int 1 d x$ $=intxe^(2-x)dx-2intx^2dx+2int1dx$

$= \int x e^{2 - x} d x - \frac{2}{3} x^{3} + 2 x$ $=intxe^(2-x)dx-2/3x^3+2x$

Let $X = 2 - x$ $X=2-x$

$x = 2 - X$ $x=2-X$

$d x = - 1 d X$ $dx=-1dX$

So:

$I = - \int (2 - X) e^{X} d x - \frac{2}{3} x^{3} + 2 x$ $I=-int(2-X)e^Xdx-2/3x^3+2x$

Using integration by parts:
$f (X) = (2 - X)$ $f(X)=(2-X)$ , $f'(X)=-1$ , $g'(X)=e^X$ , $g(X)=e^X$

So:

$I=-(2-X)e^X+int-e^XdX-2/3x^3+2x$

$I=(X-2)e^X-e^X-2/3x^3+2x$

$I=-xe^(2-x)-e^(2-x)-2/3x^3+2x+c$ , $c in RR$

And knowing that $f(0)=2$ ,

$c-e^2=2$

So:

$I=-(x+1)e^(2-x)-2/3x^3+2x+2+e^2$

What is f(x) = int xe^(2-x) -2 x^2 +2 dxf(x)=∫xe2−x−2x2+2dxf(x) = int xe^(2-x) -2 x^2 +2 dx if f(0 ) = 2 f(0)=2f(0 ) = 2 ?