JEE Main 2021 — Mathematics Calculus

$${\int }_{0}^{5}\frac{x+[x]}{{e}^{x-[x]}}dx=\alpha {e}^{-1}+\beta$$∫05​ex−[x]x+[x]​dx=αe−1+β

$$\Rightarrow {\int }_{0}^{1}\frac{x}{{e}^{x}}dx+{\int }_{1}^{2}\frac{x+1}{{e}^{x-1}}dx+{\int }_{2}^{3}\frac{x+2}{{e}^{x-2}}dx+{\int }_{3}^{4}\frac{x+3}{{e}^{x-3}}dx+{\int }_{4}^{5}\frac{x+4}{{e}^{x-4}}dx$$⇒∫01​exx​dx+∫12​ex−1x+1​dx+∫23​ex−2x+2​dx+∫34​ex−3x+3​dx+∫45​ex−4x+4​dx

Let $x-1=p,x-2=q,x-3=r,x-4=w$x−1=p,x−2=q,x−3=r,x−4=w

$$\Rightarrow {\int }_{0}^{1}\frac{x}{{e}^{x}}dx+{\int }_{0}^{1}\frac{p+2}{{e}^{p}}dp+{\int }_{0}^{1}\frac{q+4}{{e}^{q}}dq+{\int }_{0}^{1}\frac{r+6}{{e}^{r}}dr+{\int }_{0}^{1}\frac{w+8}{{e}^{w}}dw$$⇒∫01​exx​dx+∫01​epp+2​dp+∫01​eqq+4​dq+∫01​err+6​dr+∫01​eww+8​dw

$$\Rightarrow {\int }_{0}^{1}\frac{x}{{e}^{x}}dx+{\int }_{0}^{1}\frac{x+2}{{e}^{x}}dx+{\int }_{0}^{1}\frac{x+4}{{e}^{x}}dx+{\int }_{0}^{1}\frac{x+6}{{e}^{x}}dx+{\int }_{0}^{1}\frac{x+8}{{e}^{x}}dx$$⇒∫01​exx​dx+∫01​exx+2​dx+∫01​exx+4​dx+∫01​exx+6​dx+∫01​exx+8​dx

$$={\int }_{0}^{1}\frac{5x+20}{{e}^{x}}dx=5{\int }_{0}^{1}\frac{x+4}{{e}^{x}}dx$$=∫01​ex5x+20​dx=5∫01​exx+4​dx

$$=5{[(x+4)\frac{{e}^{-x}}{-1}]}_{0}^{1}-{\int }_{0}^{1}\frac{{e}^{-x}}{-1}dx$$=5[(x+4)−1e−x​]01​−∫01​−1e−x​dx

$=5[-5{e}^{-1}+4+(-{e}^{-1}+1)]$=5[−5e−1+4+(−e−1+1)]$=5(-6{e}^{-1}+5)=-30{e}^{-1}+25$=5(−6e−1+5)=−30e−1+25

$$\alpha =-30,\beta =25$$α=−30,β=25

$$(\alpha +\beta {)}^{2}=25$$(α+β)2=25