How to prove that $\int_0^1 e^{x^2-x} dx$ is more than than $e^{(\frac{-1}{4})}$, and less than 1 using series expansions alone? My first  instinct was to attempt to use its Taylor expansion
which is
$$ 1- x+\frac{3x^2}{2}-\frac{7x^3}{6}...$$
which allows me prove prove it's less than 1  by computing
$\int_0^1 (1- x+\frac{3x^2}{2}-\frac{7x^3}{6})dx $
however, I have no idea how to numerically verify if it's more than $e^{\frac{-1}{4}}$.
Could someone please give any advice, or guide me to a theorem which allows me to achieve the same (using expansions alone).
Note:- I'm aware that there are other methods of solving this, however, I'm interested in proving this by using expansions.
 A: Let $$I=\int_{0}^{1} e^{(x-1/2)^2} dx$$
$f(x)=e^{(x-1/2)^2}$ in [,1], attains min at $x=1/2$, so $f_{min}=1$ and max at $x=0,1$ so $f_{max}=e^{1/4}$ then we have
$$1 \le e^{(x-1/2)^2} \le e^{1/4} \implies \int _{0}^{1} 1 dx \le I \le \int_{0}^{1} e^{1/4} dx\implies  1\le I \le e^{1/4}.$$
Finally it follows that
$$e^{-1/4}\le \int_{0}^{1} e^{x^2-x} dx \le 1.$$
A: Your series expansion approach doesn't actually formally prove the upper bound because including only a finite number of terms does not tell you the amount of error.
Instead, the correct reasoning is to note that $x^2 - x \le 0$ for all $0 \le x \le 1$, since $x^2 - x = x(x-1)$ implies that for any $x \in (0,1)$, $x > 0$ but $x-1 < 0$, so their product is negative; and when $x \in \{0, 1\}$, $x^2 - x = 0$.
Hence $$e^{x^2 - x} \le e^0 = 1$$ on this interval and $$\int_{x=0}^1 e^{x^2 - x} \, dx \le 1.$$  Finally, the inequality is strict because when $x = 1/2$, $x^2 - x = -1/4 < 0$.
For the lower bound, we note $$x^2 - x = (x-1/2)^2 - 1/4,$$ and since the square of a real number is never negative, $$x^2 - x \ge -1/4.$$  Thus using the same reasoning as for the upper bound,
$$\int_{x=0}^1 e^{x^2 - x} \, dx \ge \int_{x=0}^1 e^{-1/4} \, dx = e^{-1/4},$$ and again, the inequality is strict because, for instance, when $x = 0$, $x^2 - x = 0 > -1/4$.
Your original comment, that you are trying to prove the integral is more than $e^{-0.25}$, reflects a misunderstanding of the comment you replied to:  note that the inequality provided is $$0 < \color{red}{x - x^2} < \frac{1}{4},$$ that is to say, $$-\frac{1}{4} < x^2 - x < 0.$$  The original comment's advice is correct as written.
The series approach is destined to fail for the same reason that you cannot assert, for instance,
$$\int_{x=0}^1 1 - \frac{1}{1!} + \frac{1^2}{2!} \, dx \le \int_{x=0}^1 e^{-1} \, dx.$$  The next term in the series expansion is $-1/3! < 0$.  In order to use such reasoning you must quantify the error term, and as such, it would be unnecessarily complicated compared to the solution provided above.
