Bounds on Maclaurin series of $e^{-x^2}$ This is a problem from a textbook:

By taking the 4th degree Maclaurin polynomial for $e^{-x^2}$ find an approximation to $\int^1_0 e^{-x^2} \,dx$. Place bounds on the error in this approximation.

The first part is done by substituting $-x^2$ in series
$$ e^{x}  =  1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
yielding
$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} + \cdots $$
Integrating gives
$$ \int^1_0 e^{-x^2}\,dx \approx x - \frac{x^3}{3} + \frac{x^5}{10} \bigg|^1_0 = \frac{23}{30}$$
Which is correct according the answer. However, I do not know how they have computed the bound,

error < $2.38 \times 10$

May someone explain? Thank you so much!

The reason is that your sequence 
$$1 - x^2 + \frac{x^4}{2!}- \frac{x^6}{3!} + .... = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!}$$
is alternating. In this case, if you use the first $n$-th term to approximate the value, then the error will be bounded by the $n+1$- term. (I hope this is proved in your textbook). In your case, the next term will be (forgetting the negative sign)
$$\frac{x^7}{7\cdot  3!}\bigg|_0^1 = \frac{1}{42} \sim 0.0238095....$$
