If [x] denotes the greatest integer less than | vedclass

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(B) Let $I = \int_{1}^{5} [|x - 3|] \, dx$. We split the integral based on the definition of the absolute value and the greatest integer function: $I

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$2$

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(B) Let $I = \int_{1}^{5} [|x - 3|] \, dx$. We split the integral based on the definition of the absolute value and the greatest integer function: $I = \int_{1}^{3} [-(x - 3)] \, dx + \int_{3}^{5} [x - 3] \, dx$. For the first part,$\int_{1}^{3} [3 - x] \, dx$: When $1 \le x When $2 \le x Thus,$\int_{1}^{3} [3 - x] \, dx = \int_{1}^{2} 1 \, dx + \int_{2}^{3} 0 \, dx = [x]_{1}^{2} = 2 - 1 = 1$. For the second part,$\int_{3}^{5} [x - 3] \, dx$: When $3 \le x When $4 \le x Thus,$\int_{3}^{5} [x - 3] \, dx = \int_{3}^{4} 0 \, dx + \int_{4}^{5} 1 \, dx = [x]_{4}^{5} = 5 - 4 = 1$. Therefore,$I = 1 + 1 = 2$.

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\int_{1}^{5} [|x - 3|] \, dx$ is

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