The value of $\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} }$, where [ x ] is the greatest integer $\le$ x, is :
Solution
$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx}$<br><br>Here, $n - 1 \le x < n$<br><br>$\therefore$ $[x] = n - 1$<br><br>$\therefore$ $\int\limits_{n - 1}^n {{e^{x - (n - 1)}}} dx$<br><br>$= \left[ {{e^{x - (n - 1)}}} \right]_{n - 1}^n$<br><br>$= {e^1} - {e^0}$<br><br>$= e - 1$<br><br>Now, $\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)}$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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