Evaluate the definite integral int_0^3 [x] , | vedclass

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(A) The integral is defined as $\int_0^3 [x] \, dx$. Since $[x]$ changes its value at every integer,we split the integral into intervals: $\int_0^3 [x

Vedclass · Accepted Answer

$3$

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(A) The integral is defined as $\int_0^3 [x] \, dx$. Since $[x]$ changes its value at every integer,we split the integral into intervals: $\int_0^3 [x] \, dx = \int_0^1 [x] \, dx + \int_1^2 [x] \, dx + \int_2^3 [x] \, dx$ In the interval $[0, 1)$,$[x] = 0$. In the interval $[1, 2)$,$[x] = 1$. In the interval $[2, 3)$,$[x] = 2$. Substituting these values: $= \int_0^1 0 \, dx + \int_1^2 1 \, dx + \int_2^3 2 \, dx$ $= 0 + [x]_1^2 + [2x]_2^3$ $= (2 - 1) + (6 - 4)$ $= 1 + 2 = 3$.

Evaluate the definite integral $\int_0^3 [x] \, dx$,where $[x]$ denotes the greatest integer function.

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