QJEEM21Ma73e2: JEE Main 2021 Mathematics - Class 12 JEEM21Ma73e2 Solution: If the value of the integral , where , R, 5 + 6 = 0, and [x]…

$I = \int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta }$

$I = \int\limits_0^1 {{x \over {{e^x}}}dx + \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx + \int\limits_2^3 {{{x + 2} \over {{e^{x - 2}}}}dx + \int\limits_3^4 {{{x + 3} \over {{e^{x - 3}}}}dx + \int\limits_4^5 {{{x + 4} \over {{e^{x - 4}}}}dx} } } } }$

Let $I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5}$

Here, ${I_2} = \int\limits_1^2 {{{x + 1} \over {{e^{x - 1}}}}dx}$ Put $x = t + 1 \Rightarrow dx = dt$

$= \int\limits_0^1 {{{t + 2} \over {{e^t}}}dt = \int\limits_0^1 {{t \over {{e^t}}}dt + \int\limits_0^1 {{2 \over {{e^t}}}dt} } }$

${I_2} = {I_1} + 2\int\limits_0^1 {{e^{ - t}}dt = {I_1} + 2(1 - {e^{ - 1}})}$

Similarly,

${I_3} = {I_1} + 4(1 - {e^{ - 1}})$

${I_4} = {I_1} + 6(1 - {e^{ - 1}})$

${I_5} = {I_1} + 8(1 - {e^{ - 1}})$

$I = {I_1} + {I_2} + {I_3} + {I_4} + {I_5} = 5{I_1} + (2 + 4 + 6 + 8)(1 - {e^{ - 1}})$

$= 5{I_1} + 20(1 - {e^{ - 1}})$

${I_1} = \int_0^1 {x{e^{ - 1}}dx = - [{e^{ - x}}(x + 1)_0^1 = 1 - 2{e^{ - 1}}}$

$\therefore$ $5{I_1} + 20(1 - {e^{ - 1}}) = 5(1 - 2{e^{ - 1}}) + 20(1 - {e^{ - 1}}) = 25 - 30{e^{ - 1}}$

$\therefore$ $\alpha$ = $-$ 30, $\beta$ = 25

Also it satisfy $5\alpha + 6\beta = 0$

Now, ${(\alpha + \beta )^2} = {( - 30 + 25)^2} = {( - 5)^2} = 25$