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)
$$\frac{x^7}{7\cdot 3!}\bigg|_0^1 = \frac{1}{42} \sim 0.0238095....$$