JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 7)
Consider the integral
$$I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $$,
where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
$$I = \int_0^{10} {{{[x]{e^{[x]}}} \over {{e^{x - 1}}}}dx} $$,
where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
45 (e $$-$$ 1)
45 (e + 1)
9 (e + 1)
9 (e $$-$$ 1)
Explanation
$$I = \int_0^{10} {[x]\,.\,{e^{[x] + 1 - x}}} dx$$
$$ = \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx} $$
$$ = - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\} $$
$$ = 45(e - 1)$$
$$ = \int_0^1 {0\,dx} + \int_1^2 {{e^{2 - x}}dx + \int_2^3 {2\,.\,{e^{3 - x}}dx} + \int_3^4 {3.{e^{4 - x}}dx} } + ......... + \int_9^{10} {9\,{e^{10 - x}}dx} $$
$$ = - \{ (1 - e) + 2(1 - e) + 3(1 - e) + ....... + 9(1 - e)\} $$
$$ = 45(e - 1)$$
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