$I(n)=∫_1^n(1+2^{-[x] })^x dx $ Prove: $n-11$ Let 
$ I(n)=∫_1^n(1+2^{-[x] })^x dx $ Prove: $n-1&lt;I(n)&lt;n+1,nϵN,n&gt;1$.


  
    To solve this problem I broke the iintegral and find each integral,looking for q better solu
$\newcommand{\d}[1]{\text{ d}#1}$
Before I present my solutions, I would first highlight that your inequality on the left is false for $n = 2$, as:
$$
I(2) = \int_1^2 (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \int_1^2 \left(1 + \frac{1}{2}\right)^x \d{x} \approx 1.84973 &lt; 2
$$
However, it is true that $n - 1 &lt; I(n) &lt; n + 1$.



This is probably what you've done so far:
$$
I(n) = \int_1^n (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \sum_{k=1}^{n-1} \int_{k}^{k+1} (1 + 2^{-\lfloor{x}\rfloor})^x \d{x} = \sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x}
$$
Observe that for any $a &gt; 1$, $a^x$ is strictly convex. For any convex function $f : [c,d] \to \mathbb{R}$,  we have $f$ lying below the line joining $(c, f(c))$ and $(d, f(d))$. This means that the area under $f$ is $\leq$ the area of the trapezium with the vertices $(c,0)$, $(c, f(c))$, $(d, f(d))$ and $(d,0)$. Thus:
\begin{align*}
\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} &amp;&lt; \frac{1}{2}((1+2^{-k})^k + (1+2^{-k})^{k+1}) \\
&amp;= \frac{1}{2}(1+2^{-k})(2 + 2^{-k}) \\
&amp;= 1 + 2^{-k} + 2^{-k-1} + 2^{-2k-1}
\end{align*}
Therefore:
\begin{align*}
\sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} &amp;&lt; \sum_{k=1}^{n-1}(1 + 2^{-k} + 2^{-k-1} + 2^{-2k-1}) \\
&amp;= n - \frac{4^{1-n}}{3} - 3(2^{-n}) + \frac{5}{6} \\
&amp;&lt; n + 1
\end{align*}
To show that $I(n) &gt; n - 1$, simply observe that $(1 + 2^{-k})^x &gt; 1$ for all $k \geq 1$, so:
$$
I(n) = \sum_{k=1}^{n-1}\int_{k}^{k+1} (1 + 2^{-k})^x \d{x} &gt; \sum_{k=1}^{n-1}\int_{k}^{k+1} 1 \d{x} = n - 1 
$$
Split your I tegral up between when floor(x) changes. So we get sum_k=1 to n-1 of integral from k to k+1 (1+2^-k)^x dx.

The integral will be 1/ln(1+2^-k)*[(1+2^-k)^k+1 - (1+2^-k)^k] = 2^-k * (1+2^-k)^(k+1) /ln(1+2^-k).

I don't know how you can show that the sum of these integrals will be between n a d n+1, so I will stop here.