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Prove the following integral inequality:

$$ 1.462 \le \int_0^1 e^{{x}^{2}}\le 1.463$$

This is a high school problem. So far i did manage to prove that the integral is bigger than $1.462$ by using Taylor expansion, namely: $$1.462\le 1.4625=\int_0^1 1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+\frac{x^{10}}{120}\le \int_0^1 e^{{x}^{2}}$$ For the right bound i'm still looking for a way. However, i wonder if there is an elegant way to solve both sides.

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Taylor out a few more terms and use the taylor error term for the uppers bound, perhaps? – mixedmath Jul 18 '12 at 8:33
@mixedmath: it could be. – user 1618033 Jul 18 '12 at 8:35
I notice you refer to yourself as "i". Do you consider yourself to be imaginary? :) – Tom Zych Jun 16 '14 at 10:30
up vote 3 down vote accepted

Here is how to find the upper bound, integration by parts gives

$$ \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx}={{\rm e}}-\int _{0}^{1}\!2 \,{x}^{2}{{\rm e}^{{x}^{2}}}{dx}$$

Using the fact that

$$ 2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{10}+{\frac {1}{ 60}}\,{x}^{12}\leq 2\,{x}^{2}{{\rm e}^{{x}^{2}}}$$


$$ \int _{0}^{1} (\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\leq \int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}\,,$$

since both functions are positive. Multiplying both sides of the above inequality by -1, yields,

$$-\int _{0}^{1}(\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\geq -\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}$$

adding e to both sides of the last inequality gives

$$ e-\int _{0}^{1}(\!2\,{x}^{2}+2\,{x}^{4}+{x}^{6}+1/3\,{x}^{8}+1/12\,{x}^{ 10}+{\frac {1}{60}}\,{x}^{12}){dx}\geq e-\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx}$$

Evaluating the integral of the approximate power series gives the upper bound

$$ \int _{0}^{1}\!{{\rm e}^{{x}^{2}}}{dx} = e-\int _{0}^{1}\!2\,{x}^{2}{ {\rm e}^{{x}^{2}}}{dx} \leq 1.462863173 < 1.463 $$

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@Mhenni Benghorbal: now it's much better! :-) – user 1618033 Jul 19 '12 at 9:07
@Mhenni Benghorbal: maybe you take some interest in…. I'm curious about a nice simple proof for the right side of the inequality. – user 1618033 Jul 19 '12 at 9:09
@Chris's sister: I'll give it a try. – Mhenni Benghorbal Jul 19 '12 at 10:55
@MhenniBenghorbal: +1 Fine! – Raymond Manzoni Jul 19 '12 at 11:24

You can use Taylor's theorem with error. You are on the right track.

Suppose we are creating a Taylor polynomial about $a$. There is a theorem that states that if, for fixed numbers $$m \le f^{n+1}(t) \le M$$ for all $t$ on some interval containing $a$, then the error $E(x)$ satisfies the inequalities $$\frac{m(x-a)^n}{(n+1)!} \le E(x) \le \frac{M(x-a)^n}{(n+1)!}$$ for $x>a$. (This is theorem 7.7 in Apostol's Calculus Vol. 1)

For our problem, we are using a Taylor polynomial centered around $0$, so $a=0$. Take the interval $[0,1]$. Then, we have the bounds $0 \le f^{2n}(t) \le \frac{(2n)!}{n!}$.

Substituting $2n$ for $n+1$ and $2n-1$ for $n$, we get $$0 \le E(x) \le \frac{x^{2n-1}}{n!}$$

Integrating the error, we get $$0 \le \int_0^1 E(x) dx \le \frac{1}{n!\cdot 2n}$$

It is important to note that $n$ is $(d-1) / 2$ where $d$ denotes the degree of the polynomial, because we substituted $2n$ instead of $n+1$.

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Integrating the power series for $e^{x^2}$, term by term, gives $$ \int_0^1e^{x^2}\,\mathrm{d}x=\sum_{k=0}^\infty\frac1{(2k+1)k!} $$ Since $$ \begin{align} \sum_{k=n+1}^\infty\frac1{(2k+1)k!} &\le\frac1{(2n+3)(n+1)!}\left[1+\frac1{n+2}+\frac1{(n+2)^2}+\dots\right]\\ &=\frac1{(2n+3)(n+1)!}\frac{n+2}{n+1}\\ &\le\frac1{(2n+1)(n+1)!} \end{align} $$ If we use $n=5$ in $$ \sum_{k=0}^n\frac1{(2k+1)k!}\le\int_0^1e^{x^2}\,\mathrm{d}x\le\frac1{(2n+1)(n+1)!}+\sum_{k=0}^n\frac1{(2k+1)k!} $$ we get $$ 1.4625300625\le\int_0^1e^{x^2}\,\mathrm{d}x\le1.4626563252 $$

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