The value of int_{0.2}^{3.5} [x] , dx is equa | vedclass

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(B) We are given the definite integral $\int_{0.2}^{3.5} [x] \, dx$,where $[x]$ denotes the greatest integer function.We split the interval $[0.2, 3.5

Vedclass · Accepted Answer

$4.5$

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(B) We are given the definite integral $\int_{0.2}^{3.5} [x] \, dx$,where $[x]$ denotes the greatest integer function.We split the interval $[0.2, 3.5]$ based on the points where the greatest integer function changes its value:$\int_{0.2}^{3.5} [x] \, dx = \int_{0.2}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx$Since $[x] = 0$ for $x \in [0.2, 1)$,$[x] = 1$ for $x \in [1, 2)$,$[x] = 2$ for $x \in [2, 3)$,and $[x] = 3$ for $x \in [3, 3.5]$,we have:$= \int_{0.2}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{3.5} 3 \, dx$$= 0 + [x]_{1}^{2} + 2[x]_{2}^{3} + 3[x]_{3}^{3.5}$$= 0 + (2 - 1) + 2(3 - 2) + 3(3.5 - 3)$$= 0 + 1 + 2(1) + 3(0.5)$$= 1 + 2 + 1.5 = 4.5$

The value of $\int_{0.2}^{3.5} [x] \, dx$ is equal to:

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