Answer
$x,x^{2},\frac{x^{3}}{2}, \frac{x^{4}}{6}$
Work Step by Step
$f(x)=xe^{x}$
$f'(x)=(x+1)e^{x}; f'(0)=1$
$f''(x)=(x+2)e^{x}; f''(0)=2$
$f'''(x)=(x+3)e^{x}; f'''(0)=3$
$f^{(4)}(x)=(x+4)e^{x}; f^{(4)}(0)=4$
n-th term of Taylor series is $\dfrac{f^{n-1}(0)x^{n-1}}{(n-1)!}$
Ist term of Taylor series is $\frac{f(0)x^{0}}{(0)!}=0$
Second term of Taylor series is $\frac{f'(0)x^{1}}{1!}=x$
Third term of Taylor series is $\frac{f''(0)x^{2}}{2!}=x^{2}$
Fourth term of Taylor series is $\frac{f'''(0)x^{3}}{3!}=\frac{x^{3}}{2}$
Fifth term of Taylor series is $\frac{f^{(4)}(0)x^{4}}{4!}=\frac{x^{4}}{6}$
Hence the first four non-zero terms of the Taylor's series are:
$x, x^{2}, \frac{x^{3}}{2}, \frac{x^{4}}{6}$
