#### Answer

$$T_4=x-x^3$$

#### Work Step by Step

Given $$f(x)=x e^{-x^{2}}, \quad T_{4}, \quad a=0$$
Since
\begin{array}{rlrl}
{f(x)} & {=x e^{-x^{2}}} & {f(0)} & {=0} \\
{f^{\prime}(x)} & {=-\left(2 x^{2}-1\right) e^{-x^{2}}} & {f^{\prime}(0)} & {=1} \\
{f^{\prime \prime}(x)} & {=\left(4 x^{3}-6 x\right) e^{-x^{2}}} & {f^{\prime \prime}(0)} & {=0} \\
{f^{\prime \prime \prime}(x)} & {=-\left(8 x^{4}-24 x^{2}+6\right) e^{-x^{2}}} & {f^{\prime \prime \prime}(0)} & {=-6} \\
{f^{(4)}(x)} & {=\left(16 x^{5}-80 x^{3}+60 x\right) e^{-x^{2}}} & {f^{(4)}(0)} & {=0}
\end{array}
Then
\begin{aligned}
T_{4}(x)&= f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}++\frac{f^{(4)}(a)}{4 !}(x-a)^{4} \\
&= x-\frac{6}{3!}x^3=x-x^3
\end{aligned}