JEE Main 2021 — Definite Integration Question with Solution

$$I=\underset{0}{\overset{5}{\int }}\frac{x+[x]}{e^{x-[x]}}dx=\alpha e^{-1}+\beta$$

$$I=\underset{0}{\overset{1}{\int }}\frac{x}{e^x}dx+\underset{1}{\overset{2}{\int }}\frac{x+1}{e^{x-1}}dx+\underset{2}{\overset{3}{\int }}\frac{x+2}{e^{x-2}}dx+\underset{3}{\overset{4}{\int }}\frac{x+3}{e^{x-3}}dx+\underset{4}{\overset{5}{\int }}\frac{x+4}{e^{x-4}}dx$$

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

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

$$=\underset{0}{\overset{1}{\int }}\frac{t+2}{e^t}dt=\underset{0}{\overset{1}{\int }}\frac{t}{e^t}dt+\underset{0}{\overset{1}{\int }}\frac{2}{e^t}dt$$

$$I_2=I_1+2\underset{0}{\overset{1}{\int }}{e}^{-t}dt=I_1+2(1-e^{-1})$$

Similarly,

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

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

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

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

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

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

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

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

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

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