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

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Answer 1 · 2018-07-30T15:49:50

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

$f (x) = 1 + \int_{- 1}^{x} (t e^{t} - t) d t$ $f(x) = 1+ int_(-1)^x (te^t-t)dt$

as the integral is linear:

$f (x) = 1 + \int_{- 1}^{x} t e^{t} d t - \int_{- 1}^{x} t d t$ $f(x) = 1+ int_(-1)^x te^tdt- int_(-1)^x tdt$

Solve the two integrals separately:

$\int_{- 1}^{x} t d t = {[\frac{t^{2}}{2}]}_{- 1}^{x}$ $int_(-1)^x tdt = [t^2/2]_(-1)^x$

$\int_{- 1}^{x} t d t = \frac{x^{2}}{2} - \frac{1}{2}$ $int_(-1)^x tdt = x^2/2-1/2$

and:

$\int_{- 1}^{x} t e^{t} d t = \int_{- 1}^{x} t \frac{d}{d t} (e^{t}) d t$ $int_(-1)^x te^tdt = int_(-1)^x t d/dt(e^t) dt$

integrating by parts:

$\int_{- 1}^{x} t e^{t} d t = - \int_{- 1}^{x} e^{t} d t + {[t e^{t}]}_{- 1}^{t}$ $int_(-1)^x te^tdt = - int_(-1)^x e^t dt + [te^t]_( -1)^x$

$\int_{- 1}^{x} t e^{t} d t = - {[e^{t}]}_{- 1}^{t} + x e^{x} + \frac{1}{e}$ $int_(-1)^x te^tdt = - [e^t]_(-1)^x + xe^x+1/e$

$\int_{- 1}^{x} t e^{t} d t = - e^{x} + \frac{1}{e} + x e^{x} + \frac{1}{e}$ $int_(-1)^x te^tdt = - e^x+1/e + xe^x+1/e$

$\int_{- 1}^{x} t e^{t} d t = e^{x} (x - 1) + \frac{2}{e}$ $int_(-1)^x te^tdt = e^x(x-1)+2/e$

Putting together the partial results:

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

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

What is f(x) = int xe^x-x dxf(x)=∫xex−xdxf(x) = int xe^x-x dx if f(-1) = 1 f(−1)=1f(-1) = 1 ?