The value of int_1^4 log [x] dx,where [x] is | vedclass

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(B) We need to evaluate the integral $I = \int_1^4 \log [x] dx$. Since $[x]$ is the greatest integer function,we split the integral at the integer poi

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$\log 6$

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(B) We need to evaluate the integral $I = \int_1^4 \log [x] dx$. Since $[x]$ is the greatest integer function,we split the integral at the integer points $x=2$ and $x=3$: $I = \int_1^2 \log [x] dx + \int_2^3 \log [x] dx + \int_3^4 \log [x] dx$. For $x \in [1, 2)$,$[x] = 1$,so $\log [x] = \log 1 = 0$. For $x \in [2, 3)$,$[x] = 2$,so $\log [x] = \log 2$. For $x \in [3, 4)$,$[x] = 3$,so $\log [x] = \log 3$. Substituting these values: $I = \int_1^2 0 dx + \int_2^3 \log 2 dx + \int_3^4 \log 3 dx$. $I = 0 + [x \log 2]_2^3 + [x \log 3]_3^4$. $I = (3-2) \log 2 + (4-3) \log 3$. $I = 1 \cdot \log 2 + 1 \cdot \log 3 = \log 2 + \log 3$. Using the property $\log a + \log b = \log(ab)$,we get $I = \log(2 	imes 3) = \log 6$.

The value of $\int_1^4 \log [x] dx$,where $[x]$ is the greatest integer function less than or equal to $x$,is equal to:

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