int_{0.2}^{3.5} [x] , dx = (where [x] is the | vedclass

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(C) We evaluate the definite integral by splitting the interval based on the jumps of the greatest integer function $[x]$: $\int_{0.2}^{3.5} [x] \, dx

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(C) We evaluate the definite integral by splitting the interval based on the jumps of the greatest integer function $[x]$: $\int_{0.2}^{3.5} [x] \, dx = \int_{0.2}^{1} [x] \, dx + \int_{1}^{2} [x] \, dx + \int_{2}^{3} [x] \, dx + \int_{3}^{3.5} [x] \, dx$ 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 $x \in [3, 3.5)$: $= \int_{0.2}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{3.5} 3 \, dx$ $= 0 + [x]_{1}^{2} + 2[x]_{2}^{3} + 3[x]_{3}^{3.5}$ $= 0 + (2 - 1) + 2(3 - 2) + 3(3.5 - 3)$ $= 0 + 1 + 2(1) + 3(0.5)$ $= 1 + 2 + 1.5 = 4.5$

$\int_{0.2}^{3.5} [x] \, dx = $ (where $[x]$ is the greatest integer function not greater than $x$)

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