If the value of the integral _0^5x+[x]e^x-[x],dx= e^-1+, where ,R, 5+6=0 and [x] denotes t…

Question

If the value of the integral $\displaystyle\int_0^5\frac{x+[x]}{e^{x-[x]}}\,dx=\alpha e^{-1}+\beta$, where $\alpha,\beta\in\mathbb{R}$, $5\alpha+6\beta=0$ and $[x]$ denotes the greatest integer less than or equal to $x$, then the value of $(\alpha+\beta)^2$ is equal to:

Accepted Answer

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

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

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

Solutions · 1