The value of _n=1^100_n-1^n e^x-[x],dx, where [x] is the greatest integer x, is

Question

The value of $\displaystyle\sum_{n=1}^{100}\int_{n-1}^n e^{x-[x]}\,dx$, where $[x]$ is the greatest integer $\le x$, is

Accepted Answer

$$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx} $$

Here, $$n - 1 \le x < n$$

$$ \therefore $$ $$[x] = n - 1$$

$$ \therefore $$ $$\int\limits_{n - 1}^n {{e^{x - (n - 1)}}} dx$$

$$ = \left[ {{e^{x - (n - 1)}}} \right]_{n - 1}^n$$

$= {e^1} - {e^0}$

$= e - 1$

Now, $$\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)} $$

Solutions · 1