Consider the integral I = int_{0}^{10} frac{[ | vedclass

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(C) The integral is given by $I = \int_{0}^{10} [x] e^{[x]-x+1} dx$. Since $[x] = n$ for $x \in [n, n+1)$,we can split the integral as: $I = \sum_{n=0

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$45(e-1)$

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(C) The integral is given by $I = \int_{0}^{10} [x] e^{[x]-x+1} dx$. Since $[x] = n$ for $x \in [n, n+1)$,we can split the integral as: $I = \sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n-x+1} dx$. Evaluating the integral for each term: $I = \sum_{n=0}^{9} n \left[ -e^{n-x+1} ight]_{n}^{n+1} = \sum_{n=0}^{9} n \left( -e^{0} + e^{1} ight) = (e-1) \sum_{n=0}^{9} n$. Using the sum formula $\sum_{n=0}^{9} n = \frac{9 	imes 10}{2} = 45$,we get: $I = 45(e-1)$.

Consider the integral $I = \int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} dx$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

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