Answer
$f(x)=\displaystyle \frac{1}{2}e^{x^{2}-2x}+\frac{1}{2}$
Work Step by Step
Slope of tangent line:
$f'(x)=(x-1)e^{x^{2}-2x}$
so,
$f(x)=\displaystyle \int(x-1)e^{x^{2}-2x}dx=$
Substitute:
$\left[\begin{array}{ll}
u=x^{2}-2x & du=(2x-2)dx\\
& du=2(x-1)dx\\
& (x-1)dx=du/2
\end{array}\right]$
$f(x)=\displaystyle \frac{1}{2}\int e^{u}du$
$=\displaystyle \frac{1}{2}e^{u}+C \qquad$
bring the variable $x$ back
$=\displaystyle \frac{1}{2}e^{x^{2}-2x}+C $
Given that $f(2)=1$, we find $C:$
$\displaystyle \frac{1}{2}e^{2^{2}-2(2)}+C =1$
$\displaystyle \frac{1}{2}\cdot e^{0}+C=1$
$C=1-\displaystyle \frac{1}{2}=\frac{1}{2}$
Thus,
$f(x)=\displaystyle \frac{1}{2}e^{x^{2}-2x}+\frac{1}{2}$
