Estimate $\int^1_0 e^{-x^2}\, dx$ Estimate $\int^1_0 e^{-x^2}\, dx$
This is in a section on Taylor series so I would assume that is how it should be solved.
I started by using the Taylor series formula for $e^x$ replacing $x$ with $x^2$. I ended up with this:
$1+(-x^2)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\frac{(-x^2)^4}{4!}+...$
It looks to me like from here I should get some Fundamental Theorem of Calculus going, but I'm not sure how. The integral evaluated at $0$ is just $1$, however the integral evaluated at $1$ is not so easy. Is there some summation formula that would apply here that I could use? 
 A: The exponential function is an entire function, hence:
$$ e^x = \sum_{n\geq 0}\frac{x^n}{n!} \tag{1} $$
leads to:
$$ \forall x\in[0,1],\quad e^{-x^2}=\sum_{n\geq 0}\frac{(-1)^n}{n!}\,x^{2n}\tag{2} $$
and by integrating termwise $(2)$ over $[0,1]$ we get:
$$ \int_{0}^{1}e^{-x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,n!}\tag{3} $$
where the RHS of $(3)$ is a series that converges pretty fast; for instance: 
$$ \left|\sum_{n\geq N}\frac{(-1)^n}{(2n+1)n!}\right|\leq\frac{1}{(2N+1)N!}\tag{4} $$
and:
$$ \int_{0}^{1}e^{-x^2}\,dx = \sum_{n=0}^{6}\frac{(-1)^n}{(2n+1)n!}+\theta = \frac{1614779}{2162160}+\theta = 0.7468360\ldots+\theta\tag{5}$$
where $|\theta|\leq\frac{1}{15\cdot 7!}$ gives:
$$ \int_{0}^{1}e^{-x^2}\,dx = \color{red}{0.7468\ldots}\tag{6}$$
Other (tight) approximations are possible by using continued fractions, since
$$ \int_{0}^{1}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,\operatorname{Erf}(1)\approx\color{blue}{\frac{3969}{1955 e}}\tag{7}$$
where the Gauss continued fraction (look for "error function" here) provides good bounds.
A: $\bf{My\;  Solution}$ We Know that in $$x\in (0,1)\;\;, x^2<x\Rightarrow -x^2>-x\Rightarrow e^{-x^2}>e^{-x}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1}e^{-x}dx = \left(1-\frac{1}{e}\right)$$
and in $$\displaystyle x\in (0,1)\;, x^2>0\Rightarrow e^{-x^2}<e^{-0}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx<\int_{0}^{1}e^{-0}dx = 1$$ 
So $$\displaystyle \left(1-\frac{1}{e}\right)< \int_{0}^{1}e^{-x^2}dx < 1$$
A: Why not use a numerical method such as Simpson's, Trapezoidal or Midpoint?
Using Taylor or a Series also work as you see above. 
A very crude option is to realise that you can calculate
$$\int_0^1e^{-ax}\,dx,$$
for a constant $a$. Therefore approximate $x^2$ by some $a$ on $[0,1]$. I found the minimum of
$$\int_0^1(x^2-ax)^2\,dx$$
at $a=\frac34$ and this yields
$$\int_0^1e^{-x^2}\,dx\approx\int_0^1e^{-\frac{3}{4}x}\,dx=\frac43\left(1-e^{-3/4}\right)$$
A: Use the standard normal distribution table.
Consider $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx$$
Let $u=\frac{x}{\sqrt{2}}$. Then $\frac{du}{dx}=\frac{1}{\sqrt{2}}$.
Then $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx=\frac{1}{\sqrt{2π}}\int e^{-u^2}\,\sqrt{2}du=\frac{1}{\sqrt{π}}\int e^{-u^2}\,du$$
So
$$\frac{1}{\sqrt{π}}\int_{0}^1 e^{-u^2}\,du=\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx$$
$$\int_{0}^1 e^{-u^2}\,du=\sqrt{π}*\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx\approx\sqrt{π}*0.4207\approx0.7457$$
