Finding upper and lower bounds for a definite integral Is there a way to find upper and lower bounds for a definite integral without explicitly finding the antiderivative?  For example, $$A\le\int_0^1\frac{1}{\sqrt{1+x^4}}dx\le B$$
I was thinking the mean value theorem for integrals here, but after working on it for a while, I've decided I don't think that will help.  Then, I was looking at somehow applying Darboux's Theorem but then I got stuck.  Any tips/help appreciated. Thanks.
 A: A classical method, going back to Newton himself, is to expand the integrand into a power series and integrate term by term.  If the resulting series converges, it converges to the value of the integral (this is easy to show if the interval of integration is within the open interval of convergence; in your case, since $1$ is right at the edge of convergence,  one needs Abel's theorem to justify the computation). If the series for the integral happens to be alternating (which is the case here), its partial sums give upper and lower bounds with any precision you want. 
So, let's use the binomial series: 
$$(1+x^4)^{-1/2} = 1-\frac12 x^4 +\frac{(-1/2)(-1/2-1)}{2!} x^8 + \frac{(-1/2)(-1/2-1)(-1/2-2)}{3!} x^8+\dots$$
for $|x|<1$. In $\sum$ notation, 
$$(1+x^4)^{-1/2} = \sum_{k=0}^\infty \frac{(-1)^k(2k-1)!!}{2^k k!} x^{4k} \tag{1}$$
Integrate (1) term by term:
$$\int_0^1 (1+x^4)^{-1/2}\,dx = \sum_{k=0}^\infty \frac{(-1)^k(2k-1)!!}{2^k k! (4k+1)}   \tag{2}$$
The terms in (2) alternate in sign, and decrease to zero in absolute value. Hence, any two consecutive terms of the series give upper and lower bounds. E.g., the sum is between 
$$\sum_{k=0}^4 \frac{(-1)^k(2k-1)!!}{2^k k! (4k+1)}  =  \frac{396193}{424320}< 0.934  
$$ and $$\sum_{k=0}^5 \frac{(-1)^k(2k-1)!!}{2^k k! (4k+1)}  =  \frac{782441}{848640}>0.922 
$$ 
(Just for comparison, the integral evaluates numerically to $0.927\dots$.)
I emphasize that the above is a calculus-textbook approach, which is rather inefficient here due to slow convergence of (2). More sophisticated methods  for estimating integrals can be  found in numerical analysis books. What method to use depends heavily on the integral we have. E.g., our function $f(x)=(1+x^4)^{-1/2}$ is concave on $[0,1]$ (its second derivative is negative). Therefore,  the trapezoidal method gives a lower bound for the integral, while the midpoint method gives an upper bound. For the integral at hand, these methods give better estimates than power series with less work. E.g., with only $5$ subintervals we get lower bound 
$$T_5 = \frac{1}{10}(f(0)+f(1))+\frac15(f(0.2)+f(0.4)+f(0.6)+f(0.8))> 0.924$$
and upper bound
$$M_5 = \frac15(f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)) < 0.929$$
Also, the integral may be transformed and/or split before a method is applied... the strategy depends on the precision needed and the computational resources available. 
A: The upper and lower bounds of definite integrals are presented as:
$$
(b-a)\inf_{x\in [a,b]}f(x)\le\int_a^b f(x)dx\le(b-a)\sup_{x\in [a,b]}f(x).
$$
So according to your example let's find the $\inf_{x\in [0,1]} f(x)$ and $\sup_{x\in [0,1]} f(x)$ for $f(x)=\frac{1}{\sqrt{1+x^4}}$. Let us calculate $\frac{df}{dx}$
$$
f'(x)=\frac{df}{dx}=\frac{-2x^3}{(1+x^4)\sqrt{1+x^4}}.
$$
By equating $f'(x)=0$, we only get $x=0$ as critical point. By checking the bounds of the interval $f(0)=1$ and $f(1)=\frac{1}{\sqrt{2}}$, we conclude that $\inf_{x\in [0,1]} f(x)=\frac{1}{\sqrt{2}}$ and $\sup_{x\in [0,1]} f(x)=1$. Therefore
$$
\frac{1}{\sqrt{2}}\le\int_a^b f(x)dx\le1.
$$
A: Hint
$$\frac1{\sqrt{2}}\le\frac{1}{\sqrt{1+x^4}}\le1,\quad \forall\;0\le x\le1$$
A: Hint: $x^4<x^2$ for $x\in(0,1)$, and $\displaystyle\int_0^1\frac{dx}{\sqrt{1+x^2}}=\text{arcsinh }1\simeq0.88$.
