Question

If $I=\underset{0}{\overset{x}{\int }}\frac{2^t}{2^{\left[t\right]}}dt,x\in R^{+}$, where $[\cdot ]$ denotes the greatest integer function, then the value of $I$ is

Answer 1

Let $x\in [n,n+1),n\in I$
Now,
$I=\underset{0}{\overset{x}{\int }}\frac{2^t}{2^{\left[t\right]}}dt\Rightarrow I=\underset{0}{\overset{x}{\int }}{2}^{\left\{t\right\}}dt\Rightarrow I=\underset{0}{\overset{n}{\int }}{2}^{\left\{t\right\}}dt+\underset{n}{\overset{x}{\int }}{2}^{\left\{t\right\}}dt\Rightarrow I=n\underset{0}{\overset{1}{\int }}{2}^{\left\{t\right\}}dt+\underset{n}{\overset{x}{\int }}{2}^{\left\{t\right\}}dt\Rightarrow I=n\underset{0}{\overset{1}{\int }}{2}^{t}dt+\underset{n}{\overset{x}{\int }}{2}^{t-n}dt\Rightarrow I=n{\left[\frac{2^t}{\ln 2}\right]}_0^1+{\left[\frac{1}{2^n}\frac{2^t}{\ln 2}\right]}_n^x\Rightarrow I=\frac{n}{\ln 2}+\frac{1}{2^n\ln 2}(2^x-2^n)\Rightarrow I=\frac{n}{\ln 2}+\frac{1}{\ln 2}(2^{x-n}-1)\therefore I=\frac{\left[x\right]+2^{\left\{x\right\}}-1}{\ln 2}$