Boundedness Property Theorem When I try to show
$$1 \leqslant \int_0^1 {\sqrt {1 + {x^3}} }  \leqslant \frac{5}{4}$$
with the Boundedness Property, I get:
$$1 \leqslant \int_0^1 {\sqrt {1 + {x^3}} }  \leqslant \sqrt{2}$$
What have I done what I did wrong?
Indeed, the property to which I refer is:
$$\begin{gathered}
  m(b - a) \leqslant \int_a^b {f(x)}  \leqslant M(b - a)  \\
  {\rm{where:}} \quad m \leqslant f(x) \leqslant M\,\quad ,\quad \,\forall x \in [a,b]  \\ 
\end{gathered} $$
Thanks for your help.
 A: The boundedness property gives a too large above bound (but it gives the below bound). To get the inequality, apply Cauchy-Schwarz inequality:
$$\int_0^1\sqrt{1+x^3}dx\leq \sqrt{\int_0^11^2dt}\sqrt{\int_0^1(1+x^3)dx}=1+\frac14=\frac54.$$
A: We will obtain a somewhat better upper bound than $5/4$, using the result that you mention.  The method is a small modification of the one that you used.
Break up the integral as 
$$\int_0^{1/2}\sqrt{1+x^3}\,dx  + \int_{1/2}^{1}\sqrt{1+x^3}\,dx.$$
Look at the first integral.  On the interval $[0,1/2]$, we have $x^3 \le 1/8$, so $1+x^3\le 9/8$, and therefore $\sqrt{1+x^3}\le 3/(2\sqrt{2})$.   Multiply by $1/2$, the length of the interval. We conclude that the first integral is $\le 3/(4\sqrt{2})$.
Look at the second integral. On the interval $[1/2,1]$, we have $x^3\le 1$, so our function is $\le \sqrt{2}$. Thus the second integral is $\le \sqrt{2}/2$.
Add up. We get
$$\int_0^1\sqrt{1+x^3}\,dx \le \frac{3}{4\sqrt{2}}+\frac{\sqrt{2}}{2}.$$
The calculator shows that the expression on the right is approximately $1.2374369$, and in particular is less than $5/4$.  We could get a better upper bound by choosing the "breakpoint" a bit larger than $1/2$, roughly $0.61$. The reason that the breaking up idea worked is that although our function is close to $\sqrt{2}$ near $x=1$, it is quite a bit smaller than $\sqrt{2}$ on the interval $[0,1/2]$.
