If the value of the integral int_{0}^{5} frac | vedclass

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(B) Let $I = \int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx$. Since $[x]$ is the greatest integer function,we split the integral at integer points:$I = \su

Vedclass · Accepted Answer

$25$

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(B) Let $I = \int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx$. Since $[x]$ is the greatest integer function,we split the integral at integer points:$I = \sum_{k=0}^{4} \int_{k}^{k+1} \frac{x+k}{e^{x-k}} \,dx$.Let $t = x-k$,then $dx = dt$. When $x=k, t=0$ and when $x=k+1, t=1$. The integral becomes:$I = \sum_{k=0}^{4} \int_{0}^{1} \frac{t+k+k}{e^{t}} \,dt = \sum_{k=0}^{4} \int_{0}^{1} (t+2k) e^{-t} \,dt$.$I = \int_{0}^{1} (t+0)e^{-t} dt + \int_{0}^{1} (t+2)e^{-t} dt + \int_{0}^{1} (t+4)e^{-t} dt + \int_{0}^{1} (t+6)e^{-t} dt + \int_{0}^{1} (t+8)e^{-t} dt$.$I = \int_{0}^{1} (5t + 20)e^{-t} dt = 5 \int_{0}^{1} (t+4)e^{-t} dt$.Using integration by parts $\int (t+4)e^{-t} dt = -(t+4)e^{-t} - \int -e^{-t} dt = -(t+4)e^{-t} - e^{-t} = -(t+5)e^{-t}$.Evaluating from $0$ to $1$: $5[-(1+5)e^{-1} - (-(0+5)e^{0})] = 5[-6e^{-1} + 5] = -30e^{-1} + 25$.Comparing with $\alpha e^{-1} + \beta$,we get $\alpha = -30$ and $\beta = 25$.Check condition: $5(-30) + 6(25) = -150 + 150 = 0$. This holds true.Thus,$(\alpha + \beta)^{2} = (-30 + 25)^{2} = (-5)^{2} = 25$.

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

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