JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 7)
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 :
100e
100(e $$-$$ 1)
100(1 + e)
100(1 $$-$$ e)
Explanation
$$\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}} dx} $$
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, $$\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)} $$
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, $$\sum\limits_{n = 1}^{100} {(e - 1) = 100(e - 1)} $$
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